Problem: Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."

Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?

Input

The first line contains a single integer k (1 ≤ k ≤ 109).

Output

You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.

The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them. 

Examples

Input

2

Output

4
NNYY
NNYY
YYNN
YYNN

Input

9

Output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

Input

1

Output

2
NY
YN

Note

In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.

In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long x, long long y) { return x ? gcd(y % x, x) : y; }
const long long mod = 1e9 + 7;
inline signed rd() {
  signed ans = 0;
  char last = ' ', ch = getchar();
  while (!(ch >= '0' && ch <= '9')) last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
inline long long QPow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d = a;
  if (b == 0)
    x = 1, y = 0;
  else {
    d = exgcd(b, a % b, y, x), y -= a / b * x;
  }
  return d;
}
const long double PI = 3.141592653589793115997963468544185161590576171875;
signed bits[50];
bool g[1010][1010];
long long pos[50];
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  long long n, m, A, cf, cm, pl, pr, vl, vr, k, x, y, px, py;
  cin >> k;
  signed top = 0;
  m = k;
  while (m) bits[top++] = m & 1, m >>= 1;
  signed cur = 3;
  pos[0] = 3;
  g[1][cur++] = 1;
  for (long long i = 0; i < top - 1; i++) {
    g[cur - 1][cur] = g[cur - 1][cur + 1] = 1;
    g[cur][cur + 2] = g[cur + 1][cur + 2] = 1;
    cur += 3;
    pos[i + 1] = cur - 1;
  }
  g[cur - 1][2] = 1;
  for (long long i = top - 2; i >= 0; i--) {
    if (bits[i] == 0) continue;
    long long last = pos[i];
    for (long long j = i; j < top - 1; j++) {
      g[last][cur++] = 1;
      g[cur - 1][cur++] = 1;
      last = cur - 1;
    }
    g[cur - 1][2] = 1;
  }
  cout << --cur << endl;
  for (long long i = 1; i <= cur; i++) {
    for (long long j = 1; j <= cur; j++) {
      if (i == j)
        cout << 'N';
      else
        cout << (g[i][j] || g[j][i] ? 'Y' : 'N');
    }
    cout << '\n';
  }
  return 0;
}