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.
#include <bits/stdc++.h>
using namespace std;
int a[50];
int sz;
int mp[1005][1005];
int main() {
  int k;
  scanf("%d", &k);
  if (k == 1) {
    printf("2\n");
    printf("NY\nYN\n");
    return 0;
  }
  while (k) {
    a[sz++] = k % 2;
    k /= 2;
  }
  int cnt = 2;
  for (int i = 1; i < sz - 1; i++) {
    mp[2 + i][2 + i + 1] = mp[2 + i + 1][2 + i] = 1;
  }
  mp[2 + sz - 1][2] = mp[2][2 + sz - 1] = 1;
  cnt += sz - 1;
  for (int i = 0; i < sz; i++) {
    if (a[i] == 1) {
      if (i == 0) {
        mp[1][3] = mp[3][1] = 1;
      } else {
        mp[1][cnt + 1] = mp[cnt + 1][1] = 1;
        mp[1][cnt + 2] = mp[cnt + 2][1] = 1;
        if (i == sz - 1) {
          mp[2][cnt + i * 2 - 1] = mp[cnt + i * 2 - 1][2] = 1;
          mp[2][cnt + i * 2] = mp[cnt + i * 2][2] = 1;
        }
        for (int j = 0; j < i - 1; j++) {
          mp[cnt + 2 * (j + 1) - 1][cnt + 2 * (j + 2) - 1] =
              mp[cnt + 2 * (j + 2) - 1][cnt + 2 * (j + 1) - 1] = 1;
          mp[cnt + 2 * (j + 1) - 1][cnt + 2 * (j + 2)] =
              mp[cnt + 2 * (j + 2)][cnt + 2 * (j + 1) - 1] = 1;
          mp[cnt + 2 * (j + 1)][cnt + 2 * (j + 2) - 1] =
              mp[cnt + 2 * (j + 2) - 1][cnt + 2 * (j + 1)] = 1;
          mp[cnt + 2 * (j + 1)][cnt + 2 * (j + 2)] =
              mp[cnt + 2 * (j + 2)][cnt + 2 * (j + 1)] = 1;
        }
        for (int j = i - 1; j < sz - 2; j++) {
          if (j == i - 1) {
            mp[cnt + 2 * (i - 1) + 1][2 + i + 1] =
                mp[2 + i + 1][cnt + 2 * (i - 1) + 1] = 1;
            mp[cnt + 2 * (i - 1) + 2][2 + i + 1] =
                mp[2 + i + 1][cnt + 2 * (i - 1) + 2] = 1;
          }
        }
        cnt += i * 2;
      }
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      if (mp[i][j] == 1)
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
  return 0;
}
