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 num[10000];
int map[2000][2000];
int add(int a, int b, int c) { map[a][b] = 1, map[b][a] = 1; }
int vis[2000][2];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    memset(map, 0, sizeof(map));
    memset(vis, 0, sizeof(vis));
    int i, j;
    int len = 0;
    int ans = 0;
    memset(map, 0, sizeof(map));
    memset(num, 0, sizeof(num));
    for (i = 0; i <= 30; i++) {
      if (n & (1 << i)) {
        num[i]++;
        len = i;
      }
    }
    int index = 3;
    for (i = 0; i <= 30; i++) {
      if (i != len) continue;
      int x, y;
      if (i == 0) {
        add(1, 2, 1);
        index = 3;
        continue;
      }
      for (j = 1; j <= i; j++) {
        if (j == 1) {
          add(1, index++, 3);
          add(1, index++, 4);
          x = index - 1;
          y = index - 2;
        } else {
          add(x, index, 5);
          index++;
          add(y, index, 6);
          index++;
          add(x, index - 1, 7);
          add(y, index - 2, 8);
          x = index - 1;
          y = index - 2;
        }
        vis[j][0] = x;
        vis[j][1] = y;
      }
      if (i == len) {
        add(x, 2, 1);
        add(y, 2, 1);
        continue;
      }
    }
    for (i = 0; i <= 30; i++) {
      if (num[i] && i != len) {
        int x = 1;
        for (j = 1; j <= len - i; j++) {
          add(x, index++, 1);
          x = index - 1;
        }
        if (i == 0) {
          add(x, 2, 1);
          continue;
        }
        add(x, vis[len - i + 1][0], 1);
        add(x, vis[len - i + 1][1], 1);
      }
    }
    printf("%d\n", index - 1);
    for (i = 1; i < index; i++) {
      for (j = 1; j < index; j++) {
        if (map[i][j] == 1)
          printf("Y");
        else
          printf("N");
      }
      printf("\n");
    }
  }
  return 0;
}