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 main() {
  int k;
  while (scanf("%d", &k) == 1) {
    int G[250][250] = {};
    G[1][3] = 1;
    for (int i = 3; i < 75; i++) {
      if (i % 2 == 1) {
        G[i][i + 1] = 1;
        G[i][i + 2] = 1;
      } else {
        G[i][i + 2] = 1;
        G[i][i + 3] = 1;
      }
    }
    for (int i = 200; i < 233; i++) G[i][i + 1] = 1;
    G[233][2] = 1;
    for (int i = 0; i < 32; i++)
      if (k & (1 << i)) {
        G[(i + 2) * 2][200 + i] = 1;
      }
    printf("233\n");
    for (int i = 1; i <= 233; i++) {
      for (int j = 1; j <= 233; j++)
        printf("%c", (G[i][j] || G[j][i]) ? 'Y' : 'N');
      printf("\n");
    }
  }
  return 0;
}