#include <bits/stdc++.h>
//#pragma GCC optimize("Ofast","-funroll-loops")
//#pragma GCC target("sse4.1","sse4.2","ssse3","sse3","sse2","sse","avx2","avx","popcnt","tune=native")

using namespace std;

typedef long double ldb;
typedef long long i64;
typedef unsigned long long u64;
typedef unsigned int u32;
typedef pair<int, int> pii;

// priority_queue<int, vector<int>, greater<int>> minq;
// ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
// fflush(stdout);

const int inf = 1000000007;
const i64 prm = 998244353;
const i64 inf2 = ((i64)inf) * inf;
const i64 bone = 1;
const int maxn = 500010;

inline int read(){
  int x=0,f=0; char ch=getchar();
  while(!isdigit(ch)) f|=(ch==45),ch=getchar();
  while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
  return f?-x:x;
}

/*
   1. 请再三确认题意
   2. 对于复杂的数据结构+贪心/结论/计数题，可以考虑先写暴力，确认想法正确，再逐步优化
*/

int fa[maxn];
int finf(int u) {
  return fa[u] == u ? u : fa[u]=finf(fa[u]);
}
bool adj(int u, int v) {
  int fu = finf(u), fv = finf(v);
  if (fu == fv) {
    return true;
  }
  fa[fu] = fv;
  return false;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i) fa[i] = i;
  while (n--) {
    int u = read(), v = read();
    if (adj(u, v)) {
      puts("Y");
    } else {
      puts("N");
    }
  }
  return 0;
}