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;
const int D = 30;
int k, n;
bool tag[1010][1010];
int main() {
  scanf("%d", &k);
  n = 31;
  for (int i = 3; i <= 31; i++) tag[i][i - 1] = 1;
  for (int t = 0; t <= 29; t++)
    if (k & (1 << t)) {
      if (!t) {
        tag[31][1] = 1;
        continue;
      }
      int now = 31 - t;
      tag[now][++n] = 1;
      tag[now][++n] = 1;
      for (int i = 1; i < t; i++) {
        tag[n - 1][n + 1] = 1;
        tag[n - 1][n + 2] = 1;
        tag[n][n + 1] = 1;
        tag[n][n + 2] = 1;
        n += 2;
      }
      tag[n - 1][1] = tag[n][1] = 1;
    }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (tag[i][j] | tag[j][i])
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
  return 0;
}
