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 N, K, a[40], f[1005][1005];
int main() {
  scanf("%d", &K), N = 2;
  for (; K; K >>= 1) a[++a[0]] = K & 1;
  int p = 1;
  for (int i = a[0] - 1, s = 0; i; i--) {
    f[p][N + 1] = f[N + 1][p] = 1, f[p][N + 2] = f[N + 2][p] = 1,
             f[N + 1][N + 3] = f[N + 3][N + 1] = 1,
             f[N + 2][N + 3] = f[N + 3][N + 2] = 1, N += 3, s += 2, p = N;
    if (a[i]) {
      f[1][N + 1] = f[N + 1][1] = 1, f[N + s - 1][N] = f[N][N + s - 1] = 1;
      for (int j = 1; j <= s - 2; j++)
        f[N + j][N + j + 1] = f[N + j + 1][N + j] = 1;
      N += s - 1;
    }
  }
  f[p][2] = f[2][p] = 1, printf("%d\n", N);
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) putchar(f[i][j] ? 'Y' : 'N');
    puts("");
  }
}