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 main(void) {
  long long k;
  cin >> k;
  int len = 0;
  long long kcp = k;
  while (kcp > 1) {
    len += 2;
    kcp /= 2;
  }
  len++;
  vector<vector<char> > G(2 + len * len, vector<char>(2 + len * len, 'N'));
  int i = 1;
  while (k > 1) {
    int v = i + 1;
    if (k % 2 == 1) {
      int j;
      for (j = i; j < i + len - 1; j++) G[j][j + 1] = G[j + 1][j] = 'Y';
      G[j][0] = G[0][j] = 'Y';
      v += len - 1;
    }
    G[i][v] = G[v][i] = G[i][v + 1] = G[v + 1][i] = 'Y';
    G[v][v + 2] = G[v + 2][v] = G[v + 1][v + 2] = G[v + 2][v + 1] = 'Y';
    v += 2;
    i = v;
    k /= 2;
    len -= 2;
  }
  G[i][0] = G[0][i] = 'Y';
  cout << i + 1 << endl;
  for (int j = 0; j <= i; j++) {
    for (int k = 0; k <= i; k++) cout << G[j][k];
    cout << endl;
  }
}