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 mp[100][105];
int main() {
  int k;
  int i, j;
  while (cin >> k) {
    memset(mp, 0, sizeof(mp));
    mp[1][5] = 1;
    if (k & (1 << 30)) {
      mp[1][3] = 1;
      mp[1][4] = 1;
    }
    j = 3;
    for (i = 29; i >= 1; i--) {
      mp[j][j + 3] = 1, mp[j][j + 4] = 1;
      mp[j + 1][j + 3] = 1, mp[j + 1][j + 4] = 1;
      mp[j + 2][j + 5] = 1;
      if (k & (1 << i)) {
        mp[j + 2][j + 3] = 1;
        mp[j + 2][j + 4] = 1;
      }
      j += 3;
    }
    mp[90][2] = 1, mp[91][2] = 1;
    if (k & 1) mp[92][2] = 1;
    cout << 92 << endl;
    for (i = 1; i <= 92; i++) {
      for (j = 1; j <= 92; j++) {
        if (mp[i][j] || mp[j][i])
          cout << "Y";
        else
          cout << "N";
      }
      cout << endl;
    }
  }
  return 0;
}