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>
int l[2010][2010];
int np, sta[50];
int main(void) {
  int k, i, j;
  int p, p1, p2;
  while (scanf("%d", &k) != EOF) {
    for (i = 1; i <= 1000; i++) {
      for (j = 1; j <= 1000; j++) {
        l[i][j] = 0;
      }
    }
    np = 0;
    while (k) {
      sta[++np] = k & 1;
      k >>= 1;
    }
    p = 2;
    p++;
    l[1][p] = 1;
    l[p][1] = 1;
    for (i = 2; i <= np; i++) {
      p++;
      l[p - 1][p] = 1;
      l[p][p - 1] = 1;
    }
    p1 = p + 1;
    p2 = p + 2;
    for (i = 2; i <= np; i++) {
      l[p1][p2 + 1] = 1;
      l[p2 + 1][p1] = 1;
      l[p1][p2 + 2] = 1;
      l[p2 + 2][p1] = 1;
      l[p2][p2 + 1] = 1;
      l[p2 + 1][p2] = 1;
      l[p2][p2 + 2] = 1;
      l[p2 + 2][p2] = 1;
      p1 += 2;
      p2 += 2;
    }
    l[p1][2] = 1;
    l[2][p1] = 1;
    l[p2][2] = 1;
    l[2][p2] = 1;
    p = p2;
    for (i = 2, j = 2 + np; i <= np; i++, j--) {
      if (sta[i]) {
        l[j][p1] = 1;
        l[p1][j] = 1;
        l[j][p2] = 1;
        l[p2][j] = 1;
      }
      p1 -= 2;
      p2 -= 2;
    }
    if (sta[1]) {
      p++;
      l[2 + np][p] = 1;
      l[p][2 + np] = 1;
      l[p][2] = 1;
      l[2][p] = 1;
    }
    printf("%d\n", p);
    for (i = 1; i <= p; i++) {
      for (j = 1; j <= p; j++) {
        if (l[i][j]) {
          printf("Y");
        } else {
          printf("N");
        }
      }
      printf("\n");
    }
  }
  return 0;
}
