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;
const double eps = 1e-8;
const int INF = (1 << 30) - 1;
int K;
int M[100][100];
int main() {
  scanf("%d", &K);
  M[1][3] = M[3][1] = M[1][4] = M[4][1] = 1;
  M[2][89] = M[89][2] = 1;
  int st = 2;
  for (int i = 2; i <= 29; ++i) {
    M[st + 1][st + 4] = M[st + 4][st + 1] = 1;
    M[st + 1][st + 5] = M[st + 5][st + 1] = 1;
    M[st + 2][st + 4] = M[st + 4][st + 2] = 1;
    M[st + 2][st + 5] = M[st + 5][st + 2] = 1;
    M[st + 3][st + 6] = M[st + 6][st + 3] = 1;
    st += 3;
  }
  if (K & 1) {
    M[1][90] = M[90][1] = 1;
    M[5][90] = M[90][5] = 1;
  }
  K >>= 1;
  int cnt = 2;
  while (K) {
    if (K & 1) {
      M[cnt + 1][cnt + 3] = M[cnt + 3][cnt + 1] = 1;
      M[cnt + 2][cnt + 3] = M[cnt + 3][cnt + 2] = 1;
    }
    K >>= 1;
    cnt += 3;
  }
  printf("90\n");
  for (int i = 1; i <= 90; ++i) {
    for (int j = 1; j <= 90; ++j) {
      printf("%c", M[i][j] ? 'Y' : 'N');
    }
    puts("");
  }
  return 0;
}