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;
int mat[1001][1001];
int main() {
  int i, cur, x, l, j, k;
  char str[21];
  scanf("%s", str);
  l = strlen(str);
  cur = 3;
  for (i = 0; i < l; i++) {
    mat[cur][1] = mat[1][cur] = 1;
    int x = str[i] - '0';
    if (x) {
      for (j = i; j < l - 1; j++) {
        for (k = 0; k < 10; k++) {
          mat[k + cur + 1][cur] = mat[cur][k + cur + 1] = 1;
          mat[k + cur + 1][cur + 11] = mat[cur + 11][k + cur + 1] = 1;
        }
        cur += 11;
      }
      for (j = 0; j < x; j++) {
        mat[j + cur + 1][cur] = mat[cur][j + cur + 1] = 1;
        mat[j + cur + 1][cur + x + 1] = mat[cur + x + 1][j + cur + 1] = 1;
      }
      cur += x + 1;
      for (j = 0; j < i; j++) {
        mat[cur][cur + 1] = mat[cur + 1][cur] = 1;
        cur++;
        mat[cur][cur + 1] = mat[cur + 1][cur] = 1;
        cur++;
      }
      mat[cur][2] = mat[2][cur] = 1;
      cur++;
    }
  }
  printf("%d\n", cur);
  {}
  for (i = 1; i <= cur; i++) {
    for (j = 1; j <= cur; j++) {
      if (mat[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}