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 k;
  cin >> k;
  vector<vector<int> > adj(1000, vector<int>(1000, 0));
  vector<int> bit;
  while (k != 0) {
    bit.push_back(k % 2);
    k /= 2;
  }
  int nxt = 3;
  adj[nxt][2] = 1;
  adj[2][nxt] = 1;
  int n = bit.size();
  for (int i = n - 2; i >= 0; i--) {
    adj[nxt + 1][nxt] = 1;
    adj[nxt][nxt + 1] = 1;
    adj[nxt][nxt + 2] = 1;
    adj[nxt + 2][nxt] = 1;
    adj[nxt + 3][nxt + 1] = 1;
    adj[nxt + 1][nxt + 3] = 1;
    adj[nxt + 3][nxt + 2] = 1;
    adj[nxt + 2][nxt + 3] = 1;
    nxt = nxt + 3;
  }
  adj[nxt][1] = 1;
  adj[1][nxt] = 1;
  for (int i = n - 2; i >= 0; i--) {
    if (bit[i]) {
      int j = 3 * (n - i);
      int l = 2 * (n - i - 1);
      for (int k = 1; k <= l; k++) {
        adj[nxt + k][j] = 1;
        adj[j][nxt + k] = 1;
        j = nxt + k;
      }
      adj[j][2] = 1;
      adj[2][j] = 1;
      nxt = j;
    }
  }
  cout << nxt << "\n";
  for (int i = 1; i <= nxt; i++) {
    for (int j = 1; j <= nxt; j++) {
      if (adj[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
  return 0;
}
