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 chh;
const int N = 1005;
int n;
int g[N][N], h[35];
int main() {
  int i, j, k, p, x, y, p1, p2;
  while (~scanf("%d", &n)) {
    memset(g, 0, sizeof(g));
    p = 3;
    for (k = 30; k >= 0; k--) {
      if (n >= (1 << k)) break;
    }
    if (n % 2 == 1) {
      p1 = 1;
      for (i = 1; i <= k; i++) {
        g[p1][p] = g[p][p1] = 1;
        h[i] = p;
        p1 = p, p++;
      }
      g[p1][2] = g[2][p1] = 1;
      n--;
    } else {
      g[1][p] = g[p][1] = 1;
      g[1][p + 1] = g[p + 1][1] = 1;
      p1 = p, p2 = p + 1, p += 2;
      for (i = 2; i <= k; i++) {
        g[p1][p] = g[p][p1] = 1;
        g[p2][p] = g[p][p2] = 1;
        h[i] = p;
        p1 = p2 = p, p++;
      }
      g[p1][2] = g[2][p1] = 1;
      g[p2][2] = g[2][p2] = 1;
      n -= 2;
    }
    for (i = k; i >= 0; i--) {
      if (n < (1 << i)) continue;
      n -= (1 << i);
      p1 = p2 = 1;
      for (j = 1; j <= i; j++) {
        g[p1][p] = g[p][p1] = 1;
        g[p2][p] = g[p][p2] = 1;
        g[p1][p + 1] = g[p + 1][p1] = 1;
        g[p2][p + 1] = g[p + 1][p2] = 1;
        p1 = p, p2 = p + 1;
        p += 2;
      }
      if (i == k) {
        g[p1][2] = g[2][p1] = 1;
        g[p2][2] = g[2][p2] = 1;
      } else {
        g[p1][h[i + 1]] = g[h[i + 1]][p1] = 1;
        g[p2][h[i + 1]] = g[h[i + 1]][p2] = 1;
      }
    }
    printf("%d\n", p - 1);
    for (i = 1; i <= p - 1; i++) {
      for (j = 1; j <= p - 1; j++) {
        if (g[i][j])
          printf("Y");
        else
          printf("N");
      }
      puts("");
    }
  }
  return 0;
}