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;
bool a[2109][2109];
int main() {
  int Now = 62;
  scanf("%d", &n);
  if (n >= 536870912) {
    n -= 536870912;
    a[62][2] = a[2][62] = 1;
  }
  a[1][3] = a[3][1] = 1;
  a[1][4] = a[4][1] = 1;
  for (int i = 3; i <= 59; i += 2) {
    a[i][i + 2] = a[i + 2][i] = 1;
    a[i + 1][i + 2] = a[i + 2][i + 1] = 1;
    a[i][i + 3] = a[i + 3][i] = 1;
    a[i + 1][i + 3] = a[i + 3][i + 1] = 1;
  }
  for (int i = 0; i <= 28; i++)
    if ((1 << i) & n) {
      a[i * 2 + 4][Now + 1] = a[Now + 1][i * 2 + 4] = 1;
      for (int j = 1; j < 29 - i; j++) {
        a[Now + j][Now + j + 1] = a[Now + j + 1][Now + j] = 1;
      }
      a[2][Now + 29 - i] = a[Now + 29 - i][2] = 1;
      Now += 29 - i;
    }
  printf("%d\n", Now);
  for (int i = 1; i <= Now; i++) {
    for (int j = 1; j <= Now; j++)
      if (a[i][j])
        printf("Y");
      else
        printf("N");
    printf("\n");
  }
  return 0;
}