#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 <random>
#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; }
#define COLOR(s) ("\x1b[" s "m")

////////////////////////////////////////////////////////////////////////////////
template <unsigned M_> struct ModInt {
  static constexpr unsigned M = M_;
  unsigned x;
  constexpr ModInt() : x(0U) {}
  constexpr ModInt(unsigned x_) : x(x_ % M) {}
  constexpr ModInt(unsigned long long x_) : x(x_ % M) {}
  constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {}
  constexpr ModInt(long long x_) : x(((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {}
  ModInt &operator+=(const ModInt &a) { x = ((x += a.x) >= M) ? (x - M) : x; return *this; }
  ModInt &operator-=(const ModInt &a) { x = ((x -= a.x) >= M) ? (x + M) : x; return *this; }
  ModInt &operator*=(const ModInt &a) { x = (static_cast<unsigned long long>(x) * a.x) % M; return *this; }
  ModInt &operator/=(const ModInt &a) { return (*this *= a.inv()); }
  ModInt pow(long long e) const {
    if (e < 0) return inv().pow(-e);
    ModInt a = *this, b = 1U; for (; e; e >>= 1) { if (e & 1) b *= a; a *= a; } return b;
  }
  ModInt inv() const {
    unsigned a = M, b = x; int y = 0, z = 1;
    for (; b; ) { const unsigned q = a / b; const unsigned c = a - q * b; a = b; b = c; const int w = y - static_cast<int>(q) * z; y = z; z = w; }
    assert(a == 1U); return ModInt(y);
  }
  ModInt operator+() const { return *this; }
  ModInt operator-() const { ModInt a; a.x = x ? (M - x) : 0U; return a; }
  ModInt operator+(const ModInt &a) const { return (ModInt(*this) += a); }
  ModInt operator-(const ModInt &a) const { return (ModInt(*this) -= a); }
  ModInt operator*(const ModInt &a) const { return (ModInt(*this) *= a); }
  ModInt operator/(const ModInt &a) const { return (ModInt(*this) /= a); }
  template <class T> friend ModInt operator+(T a, const ModInt &b) { return (ModInt(a) += b); }
  template <class T> friend ModInt operator-(T a, const ModInt &b) { return (ModInt(a) -= b); }
  template <class T> friend ModInt operator*(T a, const ModInt &b) { return (ModInt(a) *= b); }
  template <class T> friend ModInt operator/(T a, const ModInt &b) { return (ModInt(a) /= b); }
  explicit operator bool() const { return x; }
  bool operator==(const ModInt &a) const { return (x == a.x); }
  bool operator!=(const ModInt &a) const { return (x != a.x); }
  friend std::ostream &operator<<(std::ostream &os, const ModInt &a) { return os << a.x; }
};
////////////////////////////////////////////////////////////////////////////////

constexpr unsigned MO = 998244353;
using Mint = ModInt<MO>;

constexpr int LIM_INV = 400'010;
Mint inv[LIM_INV], fac[LIM_INV], invFac[LIM_INV];

void prepare() {
  inv[1] = 1;
  for (int i = 2; i < LIM_INV; ++i) {
    inv[i] = -((Mint::M / i) * inv[Mint::M % i]);
  }
  fac[0] = invFac[0] = 1;
  for (int i = 1; i < LIM_INV; ++i) {
    fac[i] = fac[i - 1] * i;
    invFac[i] = invFac[i - 1] * inv[i];
  }
}
Mint binom(Int n, Int k) {
  if (n < 0) {
    if (k >= 0) {
      return ((k & 1) ? -1 : +1) * binom(-n + k - 1, k);
    } else if (n - k >= 0) {
      return (((n - k) & 1) ? -1 : +1) * binom(-k - 1, n - k);
    } else {
      return 0;
    }
  } else {
    if (0 <= k && k <= n) {
      assert(n < LIM_INV);
      return fac[n] * invFac[k] * invFac[n - k];
    } else {
      return 0;
    }
  }
}

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


/*
  0: ceil child
  1: floor child
  
  ex. N = 11 = 1011(2):
    0000
    0001
    001
    0100
    0101
    011
    1000
    1001
    101
    110
    111
  reversed:
    0000
    0001
    0010
     100
     101
     110
     111
    1000
    1001
    1010
  
  2^(L-1) <= N < 2^L
  reversed:
    [0, N - 2^(L-1))       in L bits
    [N - 2^(L-1), 2^(L-1)) in (L-1) bits
    [2^(L-1), N)           in L bits
  
  ans = \sum (1 + (longest contiguous 0))
      = N + \sum[1<=k<=L] \sum [longest >= k]
      = N + \sum[1<=k<=L] (N - \sum [longest < k])
  
  [N - 2^(L-1), 2^(L-1)) and [2^(L-1), N): same as {0,1}^(L-1)
  [0, N - 2^(L-1)):
    ex. N = 22 = 10110(2)
      00****
      0100**
      01010*
  
  f[k][n] := \sum[s \in {0,1}^n] [contiguous 0 in s < k]
           = [x^n] (1 - x^k) / (1 - 2 x + x^(k+1))
           = [x^n] (1 - x^k) \sum[d>=0] (-1)^i x^((k+1)d) / (1-2x)^(d+1)
*/

int L;
char S[200'010];

Mint slow() {
  vector<vector<Mint>> f(L + 1, vector<Mint>(L + 1, 0));
  for (int k = 1; k <= L; ++k) {
    f[k][0] = 1;
    for (int n = 1; n <= k; ++n) f[k][n] = f[k][n - 1] * 2;
    f[k][k] -= 1;
    for (int n = k + 1; n <= L; ++n) f[k][n] = f[k][n - 1] * 2 - f[k][n - (k + 1)];
  }
  Mint all = 0;
  for (int i = 0; i < L; ++i) (all *= 2) += (S[i] - '0');
  Mint ans = (L + 1) * all;
  for (int k = 1; k <= L; ++k) ans -= f[k][L - 1];
  int mx = 1, now = 1;
  for (int i = 1; i < L; ++i) {
    ++now;
    if (S[i] == '0') {
      chmax(mx, now);
    } else {
      for (int k = mx + 1; k <= L; ++k) {
        // \sum[s \in 0^now {0,1}^(L-i-1)] [longest < k]
        // {0,1}^now {0,1}^(L-i-1)
        ans -= f[k][now + (L - i - 1)];
        for (int j = 0; j < now; ++j) if (j < k) {
          // exclude those starting with 0^j 1
          ans += f[k][(now - j - 1) + (L - i - 1)];
        }
      }
      now = 0;
    }
  }
  return ans;
}

int main() {
  for (; ~scanf("%s", S); ) {
    L = strlen(S);
    
    const Mint ans = slow();
    printf("%u\n", ans.x);
  }
  return 0;
}