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 flag[1010][1010];
int m;
int n;
int main() {
  int len = 60;
  scanf("%d", &m);
  int n = 1;
  int idx = 3;
  while (m) {
    if (m & 1) {
      flag[n][idx] = flag[idx][n] = 1;
      idx++;
      for (int i = 1; i < len - 1; i++) {
        flag[idx - 1][idx] = flag[idx][idx - 1] = 1;
        idx++;
      }
      flag[idx - 1][2] = flag[2][idx - 1] = 1;
    }
    m >>= 1;
    len -= 2;
    if (m) {
      flag[n][idx] = flag[idx][n] = 1;
      idx++;
      flag[n][idx] = flag[idx][n] = 1;
      idx++;
      flag[idx - 2][idx] = flag[idx][idx - 2] = 1;
      flag[idx - 1][idx] = flag[idx][idx - 1] = 1;
      n = idx;
      idx++;
    }
  }
  printf("%d\n", idx - 1);
  for (int i = 1; i <= idx - 1; i++) {
    for (int j = 1; j <= idx - 1; j++) {
      if (i == j)
        printf("N");
      else if (flag[i][j])
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
}
