#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }


using U = unsigned __int128;
using Double = long double;
struct Solver {
  /*
    0^(not multiple of (2K+1)) ==> next 1^(>=2K+1)
    1^(not multiple of (2K+1)) ==> next 0^(>=3K+1)
    
    state
    0: empty
    [   1,  2K+1]: 0^1, ..., 0^(2K+1)
    [2K+2,  7K+2]: 0^1, ..., 0^(5K+1) (need >=3K+1)
    [7K+3,  9K+3]: 1^1, ..., 1^(2K+1)
    [9K+4, 13K+4]: 1^1, ..., 1^(4K+1) (need >=2K+1)
  */
  int K, N, V;
  inline int id0 (int k) { return (    0) + k; }
  inline int id00(int k) { return (2*K+1) + k; }
  inline int id1 (int k) { return (7*K+2) + k; }
  inline int id11(int k) { return (9*K+3) + k; }
  vector<vector<int>> trans;
  vector<vector<U>> dp;
  vector<vector<Double>> DP;
  Solver(int K_) : K(K_), N(-1), V(-1), dp{} {
    V = 13*K+5;
    trans.assign(V, vector<int>(2, -2));
    trans[0][0] = id0(1);
    trans[0][1] = id1(1);
    for (int k = 1; k <= 2*K+1; ++k) {
      trans[id0(k)][0] = id0((k == 2*K+1) ? 1 : (k + 1));
      trans[id0(k)][1] = (k % (2*K+1) != 0) ? id11(1) : id1(1);
    }
    for (int k = 1; k <= 5*K+1; ++k) {
      trans[id00(k)][0] = id00((k == 5*K+1) ? (3*K+1) : (k + 1));
      trans[id00(k)][1] = (k >= 3*K+1) ? ((k % (2*K+1) != 0) ? id11(1) : id1(1)) : -1;
    }
    for (int k = 1; k <= 2*K+1; ++k) {
      trans[id1(k)][0] = (k % (2*K+1) != 0) ? id00(1) : id0(1);
      trans[id1(k)][1] = id1((k == 2*K+1) ? 1 : (k + 1));
    }
    for (int k = 1; k <= 4*K+1; ++k) {
      trans[id11(k)][0] = (k >= 2*K+1) ? ((k % (2*K+1) != 0) ? id00(1) : id0(1)) : -1;
      trans[id11(k)][1] = id11((k == 4*K+1) ? (2*K+1) : (k + 1));
    }
    dp.emplace_back(V, 0);
    DP.emplace_back(V, 0);
    for (int u = 0; u < V; ++u) if (~trans[u][0] && ~trans[u][1]) {
      dp[0][u] = 1;
      DP[0][u] = 1;
    }
    for (int n = 1; ; ++n) {
      dp.emplace_back(V, 0);
      DP.emplace_back(V, 0);
      for (int u = 0; u < V; ++u) {
        for (int a = 0; a < 2; ++a) {
          const int v = trans[u][a];
          if (~v) {
            dp[n][u] += dp[n - 1][v];
            DP[n][u] += DP[n - 1][v];
          }
        }
      }
      const Double l = log(DP[n][0]) / log(2.0L);
      if (l >= 128) {
        N = n;
// cerr<<"K = "<<K<<": N = "<<N<<", log_2(DP[N][0]) = "<<l<<"; "<<(log(DP[N-1][trans[0][0]])/log(2.0L))<<" "<<(log(DP[N-1][trans[0][1]])/log(2.0L))<<endl;
        return;
      }
    }
  }
  string encode(U x) const {
    string s(N, '?');
    {
      int u = 0;
      for (int i = 0; i < N; ++i) {
        for (int a = 0; a < 2; ++a) {
          const int v = trans[u][a];
          if (~v) {
            if (x < dp[N - 1 - i][v]) {
              s[i] = '0' + a;
              u = v;
              break;
            } else {
              x -= dp[N - 1 - i][v];
            }
          }
        }
        assert(s[i] != '?');
      }
    }
// cerr<<s<<endl;
    return s;
  }
  U decode(const string &s) const {
    vector<int> sSum(N + 1, 0);
    vector<int> iss[2];
    for (int i = 0; i < N; ++i) {
      sSum[i + 1] = sSum[i] + (s[i] - '0');
      iss[s[i] - '0'].push_back(i);
    }
    // d = (# of 1) - sSum[n]
    static char f[1310][50][270];
    for (int i = 0; i < N; ++i) {
      for (int d = -K; d <= K; ++d) {
        fill(f[i][K + d], f[i][K + d] + V, -1);
      }
    }
    fill(f[N][K + 0], f[N][K + 0] + V, -2);
    for (int i = N; --i >= 0; ) {
      for (int d = -K; d <= K; ++d) for (int u = 0; u < V; ++u) {
        for (int a = 0; a < 2; ++a) {
          int pos = sSum[i] + d;
          if (a == 0) pos = i - pos;
          if (0 <= pos && pos < (int)iss[a].size() && abs(iss[a][pos] - i) <= K) {
            const int dd = d + a - (s[i] - '0');
            const int v = trans[u][a];
            if (abs(dd) <= K && ~v && ~f[i + 1][K + dd][v]) {
              f[i][K + d][u] = a;
              break;
            }
          }
        }
      }
    }
    assert(~f[0][K + 0][0]);
    U x = 0;
    {
      string ss(N, '?');
      int d = 0;
      int u = 0;
      for (int i = 0; i < N; ++i) {
        const int a = f[i][K + d][u];
        assert(~a);
        ss[i] = '0' + a;
        if (a == 1 && ~trans[u][0]) {
          x += dp[N - 1 - i][trans[u][0]];
        }
        d = d + a - (s[i] - '0');
        u = trans[u][a];
      }
// cerr<<ss<<endl;
    }
    return x;
  }
};


#include <string>

//you may define some global variables, but it does not work if you try to transfer any information from function pigeon_num or function send to function receive through these variables.
//you had better not use the same global variables in function send and in function receive.

/*
Taskid:         The index of the subtask. If it is subtask 2, then Taskid=2.
k:              The time threshold. Suppose the i-th pigeon taking off is the p_i-th one to land. It is guaranteed that abs(i-p_i)<=k.
return value:   The number of pigeons Little E will use.
*/
int pigeon_num(int Taskid, int k){
	//you may do some initialization for SEND function here
	
  const Solver solver(k);
  return solver.N;
  
	return 5;//change this into your code
}

/*
Taskid:         The index of the subtask. If it is subtask 2, then Taskid=2.
n:              The number of pigeons Little E will use. This equals the return value of function pigeon_num.
k:              The time threshold. Suppose the i-th pigeon taking off is the p_i-th one to land. It is guaranteed that abs(i-p_i)<=k.
msg:            The content of the message.
return value:	The order of the pigeons taking off. The length of this string must be n. return_value[i]='0' means the (i+1)-th pigeon taking off is black. return_value[i]='1' means the (i+1)-th pigeon taking off is white.
*/
std::string send(int Taskid, int n, int k, __uint128_t msg){
  const Solver solver(k);
  return solver.encode(msg);
  
	return "10101";//change this into your code
}

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

class Init{
public:
	Init(){
		//you may do some initialization for RECEIVE function here
	}
};

/*
Taskid:         The index of the subtask. If it is subtask 2, then Taskid=2.
k:              The time threshold. Suppose the i-th pigeon taking off is the p_i-th one to land. It is guaranteed that abs(i-p_i)<=k.
msg:			The order of the pigeons landing. Its length is equal to n. msg[i] denotes the color of the (i+1)-th pigeon watched landing by Little F. '0' means black and '1' means white.
return value:	The content of the message. It should be correct.
*/
__uint128_t receive(int Taskid, int k, const std::string &msg){
	const static Init init;
  
  const Solver solver(k);
  return solver.decode(msg);
  
	{//change this into your code
		__uint128_t hi = 5281683694948011861llu;
		__uint128_t lo = 3195384480471073102llu;
		return hi<<64|lo; //this value equals 97429867398990605044182047185430790478
	}
}