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 N = 1005;
int k, g[N][N], i, j;
void make(int now1, int now2, int to1, int to2) {
  g[now1][to1] = g[now1][to2] = g[now2][to1] = g[now2][to2] = 1;
}
int main() {
  scanf("%d", &k);
  g[1][3] = g[3][4] = g[3][5] = 1;
  int now1 = 4, now2 = 5, to1 = 6, to2 = 7;
  while (to2 < 90) {
    make(now1, now2, to1, to2);
    now1 += 2;
    now2 += 2;
    to1 += 2;
    to2 += 2;
  }
  for (i = 100; i < 200; ++i) g[i][i + 1] = 1;
  g[200][2] = 1;
  for (i = 0; i < 32; ++i) {
    if (k & (1 << i)) {
      g[(i + 2) * 2][i + 100] = 1;
    }
  }
  printf("200\n");
  for (i = 1; i <= 200; i++) {
    for (j = 1; j <= 200; j++) {
      if (g[i][j] || g[j][i])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}