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>
int k, cnt, i, t, last1, last2, now, j;
int f[1005][1005] = {0}, a[1005];
int main() {
  int k;
  scanf("%d", &k);
  while (k) {
    a[++cnt] = k % 2;
    k /= 2;
  }
  for (i = 1; i <= cnt / 2; i++) {
    t = a[i];
    a[i] = a[cnt - i + 1];
    a[cnt - i + 1] = t;
  }
  now = 3;
  f[1][3] = true;
  last1 = 3;
  last2 = 1;
  for (i = 2; i <= cnt; i++) {
    f[last1][++now] = true;
    f[last1][++now] = true;
    now++;
    last1 = now;
    f[now - 2][now] = true;
    f[now - 1][now] = true;
    f[last2][++now] = true;
    f[now][now + 1] = true;
    now++;
    if (a[i]) f[now - 2][now] = true;
    last2 = now;
  }
  f[last1][2] = true;
  printf("%d\n", now);
  for (i = 1; i <= now; i++) {
    for (j = 1; j <= now; j++) {
      if (f[i][j]) f[j][i] = true;
      if (f[j][i]) f[i][j] = true;
      if (f[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}