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 ans[1000][1000] = {};
int main() {
  int k, f = 2, d = 30, l[30];
  scanf("%d", &k);
  ans[0][f] = ans[f][0] = 1;
  l[29] = f++;
  for (int i = 28; i >= 0; --i, ++f) {
    ans[f - 1][f] = ans[f][f - 1] = 1;
    l[i] = f;
  }
  for (int i = 29; i; --i)
    if (k & (1 << i)) {
      ans[l[i]][f] = ans[f][l[i]] = ans[l[i]][f + 1] = ans[f + 1][l[i]] = 1;
      for (int j = 1; j < i; ++j, f += 2)
        for (int a = 0; a < 2; ++a)
          for (int b = 2; b < 4; ++b) ans[f + a][f + b] = ans[f + b][f + a] = 1;
      ans[1][f] = ans[f][1] = ans[1][f + 1] = ans[f + 1][1] = 1;
      f += 2;
    }
  if (k % 2) ans[l[0]][1] = ans[1][l[0]] = 1;
  printf("%d\n", f);
  for (int i = 0; i < f; ++i) {
    for (int j = 0; j < f; ++j) printf(ans[i][j] ? "Y" : "N");
    puts("");
  }
  return 0;
}