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;
char table[1005][1006];
int main() {
  int k;
  scanf("%d", &k);
  if (k == 1) {
    printf("2\n");
    printf("NY\n");
    printf("YN\n");
    return 0;
  }
  int num[30];
  int n = 0;
  while (k != 0) {
    num[n++] = k & 1;
    if (num[n - 1] == 1) {
      k = (k - 1) / 2;
    } else {
      k = k / 2;
    }
  }
  reverse(num, num + n);
  memset(table, 'N', sizeof(table));
  int cnt = 3;
  table[1][cnt] = 'Y';
  table[cnt][1] = 'Y';
  int old = 3;
  for (int i = 2; i <= n; i++) {
    table[old][cnt + 1] = 'Y';
    table[cnt + 1][old] = 'Y';
    table[old][cnt + 2] = 'Y';
    table[cnt + 2][old] = 'Y';
    table[cnt + 1][cnt + 3] = 'Y';
    table[cnt + 3][cnt + 1] = 'Y';
    table[cnt + 2][cnt + 3] = 'Y';
    table[cnt + 3][cnt + 2] = 'Y';
    cnt += 3;
    old = cnt;
    if (num[i - 1] == 1) {
      table[1][cnt + 1] = 'Y';
      table[cnt + 1][1] = 'Y';
      int len = 2 * i - 1;
      cnt++;
      for (int j = 1; j < len - 1; j++) {
        cnt++;
        table[cnt][cnt - 1] = 'Y';
        table[cnt - 1][cnt] = 'Y';
      }
      table[cnt][old] = 'Y';
      table[old][cnt] = 'Y';
    }
  }
  table[old][2] = 'Y';
  table[2][old] = 'Y';
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      printf("%c", table[i][j]);
    }
    printf("\n");
  }
  return 0;
}