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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, cp = 2;
  cin >> n;
  bool p[1000][1000];
  for (int i = n; i > 0;) {
    int sq = sqrt(sqrt(i));
    if (sq == 0) sq++;
    vector<int> nums(4, sq);
    int cr = sq * sq * sq * sq;
    bool change = true;
    while (change) {
      change = false;
      for (int j = 0; j < 4; j++) {
        if ((cr / nums[j]) * (nums[j] + 1) <= i) {
          cr = (cr / nums[j]) * (nums[j] + 1);
          nums[j]++;
          change = true;
        }
      }
    }
    for (int j = 0; j < nums[0]; j++) {
      p[cp + j][0] = p[0][cp + j] = 1;
    }
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < nums[j]; k++) {
        for (int l = 0; l < nums[j + 1]; l++) {
          p[cp + k][cp + nums[j] + l] = 1;
          p[cp + nums[j] + l][cp + k] = 1;
        }
      }
      cp += nums[j];
    }
    for (int j = 0; j < nums[3]; j++) {
      p[cp + j][1] = p[1][cp + j] = 1;
    }
    cp += nums[3];
    i -= cr;
  }
  cout << cp << endl;
  for (int i = 0; i < cp; i++) {
    for (int j = 0; j < cp; j++) {
      if (p[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
}
