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 K, N, i, j, Where[35];
bool A[1005][1005];
int main() {
  scanf("%d", &K);
  A[1][3] = A[3][1] = 1;
  A[1][4] = A[4][1] = 1;
  N = 4;
  Where[0] = N;
  for (i = 1; i <= 30; i++) {
    N += 2;
    A[N - 1][N - 3] = A[N - 3][N - 1] = 1;
    A[N - 1][N - 2] = A[N - 2][N - 1] = 1;
    A[N][N - 3] = A[N - 3][N] = 1;
    A[N][N - 2] = A[N - 2][N] = 1;
    Where[i] = N;
  }
  for (i = 0; (1 << i) <= K; i++)
    if ((1 << i) & K) {
      N++;
      A[Where[i]][N] = A[N][Where[i]] = 1;
      for (j = i + 1; j <= 30; j++) {
        N++;
        A[N][N - 1] = A[N - 1][N] = 1;
      }
      A[N][2] = A[2][N] = 1;
    }
  printf("%d\n", N);
  for (i = 1; i <= N; i++, printf("\n"))
    for (j = 1; j <= N; j++)
      if (A[i][j])
        printf("Y");
      else
        printf("N");
  return 0;
}