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[92][92];
int main() {
  int k, n = 92, i, j, l;
  scanf("%d", &k);
  for (i = 0; i < 29; i++) {
    int x = i * 3 + 2;
    a[x][x + 3] = a[x + 3][x] = 1;
    for (j = 1; j < 3; j++) {
      for (l = 4; l < 6; l++) {
        a[x + j][x + l] = a[x + l][x + j] = 1;
      }
    }
  }
  for (i = 0; i < 30; i++) {
    if ((k >> i) & 1)
      a[(29 - i) * 3 + 2][(29 - i) * 3 + 3] =
          a[(29 - i) * 3 + 3][(29 - i) * 3 + 2] = 1;
  }
  a[0][2] = a[2][0] = 1;
  a[1][90] = a[90][1] = a[1][91] = a[91][1] = 1;
  printf("%d\n", n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (a[i][j] == 1) {
        putchar('Y');
      } else {
        putchar('N');
      }
    }
    puts("");
  }
  return 0;
}
