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;
int k, n;
char G[2222][2222];
int main() {
  int k;
  cin >> k;
  memset(G, 'N', sizeof(G));
  int n = 2;
  int step = 0;
  while (k) {
    int c = k % 3;
    if (c) {
      for (int i = 0; i < 18; ++i) {
        if (i < step) {
          n += 3;
          if (i == 0) {
            G[1][n - 0] = G[n - 0][1] = 'Y';
            G[1][n - 1] = G[n - 1][1] = 'Y';
            G[1][n - 2] = G[n - 2][1] = 'Y';
          } else {
            G[n - 3][n - 0] = G[n - 0][n - 3] = 'Y';
            G[n - 3][n - 1] = G[n - 1][n - 3] = 'Y';
            G[n - 3][n - 2] = G[n - 2][n - 3] = 'Y';
            G[n - 4][n - 0] = G[n - 0][n - 4] = 'Y';
            G[n - 4][n - 1] = G[n - 1][n - 4] = 'Y';
            G[n - 4][n - 2] = G[n - 2][n - 4] = 'Y';
            G[n - 5][n - 0] = G[n - 0][n - 5] = 'Y';
            G[n - 5][n - 1] = G[n - 1][n - 5] = 'Y';
            G[n - 5][n - 2] = G[n - 2][n - 5] = 'Y';
          }
        } else {
          ++n;
          if (i == 0) {
            G[n][1] = G[1][n] = 'Y';
          } else if (i - 1 < step) {
            G[n][n - 1] = G[n - 1][n] = 'Y';
            G[n][n - 2] = G[n - 2][n] = 'Y';
            G[n][n - 3] = G[n - 3][n] = 'Y';
          } else {
            G[n][n - 1] = G[n - 1][n] = 'Y';
          }
        }
      }
      if (c == 1) {
        ++n;
        if (step == 18) {
          G[n][n - 1] = G[n - 1][n] = 'Y';
          G[n][n - 2] = G[n - 2][n] = 'Y';
          G[n][n - 3] = G[n - 3][n] = 'Y';
        } else {
          G[n][n - 1] = G[n - 1][n] = 'Y';
        }
        G[n][2] = G[2][n] = 'Y';
      } else {
        n += 2;
        if (step == 18) {
          G[n][n - 2] = G[n - 2][n] = 'Y';
          G[n][n - 3] = G[n - 3][n] = 'Y';
          G[n][n - 4] = G[n - 4][n] = 'Y';
          G[n - 1][n - 2] = G[n - 2][n - 1] = 'Y';
          G[n - 1][n - 3] = G[n - 3][n - 1] = 'Y';
          G[n - 1][n - 4] = G[n - 4][n - 1] = 'Y';
        } else {
          G[n][n - 2] = G[n - 2][n] = 'Y';
          G[n - 1][n - 2] = G[n - 2][n - 1] = 'Y';
        }
        G[n][2] = G[2][n] = 'Y';
        G[n - 1][2] = G[2][n - 1] = 'Y';
      }
    }
    ++step;
    k /= 3;
  }
  cout << n << endl;
  for (int i = 1; i <= n; ++i) {
    G[i][n + 1] = 0;
    puts(G[i] + 1);
  }
  return 0;
}