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 G[1002][1002];
void e(int u, int v) {
  G[u][v] = 1;
  G[v][u] = 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int cnt = 3;
  e(1, cnt);
  for (int i = 0; i < 32; i++) {
    e(cnt + 1, cnt);
    cnt++;
  }
  e(2, cnt++);
  n--;
  for (int i = 0; i < 30; i++) {
    if ((1 << i) & n) {
      e(1, cnt++);
      if (i == 0) {
        e(4, cnt - 1);
        continue;
      }
      e(1, cnt++);
      for (int j = 0; j < i - 1; j++) {
        e(cnt, cnt - 1);
        e(cnt, cnt - 2);
        cnt++;
        e(cnt, cnt - 2);
        e(cnt, cnt - 3);
        cnt++;
      }
      e(cnt - 1, i + 3);
      e(cnt - 2, i + 3);
    }
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      cout << ((G[i][j]) ? 'Y' : 'N');
    }
    cout << endl;
  }
  return 0;
}