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 = 3100;
int i, j, k, l, m, n, o, p, K, L, c[N][3], a[N][N];
int main() {
  cin >> K;
  o = K;
  while (o) {
    L++;
    o >>= 1;
  }
  n = 2;
  c[0][1] = c[0][2] = ++n;
  a[n][2] = a[2][n] = 1;
  for (i = 1; i <= L; i++) {
    c[i][1] = ++n;
    c[i][2] = ++n;
  }
  for (i = 1; i <= L; i++) {
    a[c[i][1]][c[i - 1][1]] = a[c[i - 1][1]][c[i][1]] = 1;
    a[c[i][1]][c[i - 1][2]] = a[c[i - 1][2]][c[i][1]] = 1;
    a[c[i][2]][c[i - 1][1]] = a[c[i - 1][1]][c[i][2]] = 1;
    a[c[i][2]][c[i - 1][2]] = a[c[i - 1][2]][c[i][2]] = 1;
  }
  for (i = 0; i <= L; i++) {
    if (K & 1) {
      if (i > 1) {
        n++;
        a[1][n] = a[n][1] = 1;
        for (j = 1; j <= L - i - 1; j++) {
          n++;
          a[n][n - 1] = a[n - 1][n] = 1;
        }
        a[n][c[i][1]] = a[c[i][1]][n] = 1;
        a[n][c[i][2]] = a[c[i][2]][n] = 1;
      } else if (!i) {
        n++;
        a[1][n] = a[n][1] = 1;
        for (j = 1; j <= L; j++) {
          n++;
          a[n][n - 1] = a[n - 1][n] = 1;
        }
        a[n][2] = a[2][n] = 1;
      } else if (i == 1) {
        n++;
        a[1][n] = a[n][1] = 1;
        for (j = 1; j <= L; j++) {
          n++;
          a[n][n - 1] = a[n - 1][n] = 1;
        }
        a[n][2] = a[2][n] = 1;
        n++;
        a[1][n] = a[n][1] = 1;
        for (j = 1; j <= L; j++) {
          n++;
          a[n][n - 1] = a[n - 1][n] = 1;
        }
        a[n][2] = a[2][n] = 1;
      }
    }
    K >>= 1;
  }
  printf("%d\n", n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++)
      if (a[i][j]) {
        putchar('Y');
      } else
        putchar('N');
    puts("");
  }
  return 0;
}