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;
const int maxn = 128;
int k, x, y, z, i, j;
bool g[maxn][maxn];
void G(int a, int b) { g[a][b] = g[b][a] = true; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> k;
  for (G(0, 3), G(0, 4), x = 3, y = 4, z = 5; k;
       x += 3, y += 3, z += 3, k /= 2) {
    G(x, x + 3);
    G(x, y + 3);
    G(y, y + 3);
    G(y, x + 3);
    G(z, z + 3);
    if (k & 1) G(y, z);
  }
  G(1, z);
  for (cout << 100 << endl, i = 0; i < 100; cout << endl, i++)
    for (j = 0; j < 100; j++)
      if (g[i][j])
        cout << 'Y';
      else
        cout << 'N';
  return 0;
}