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.
#include <bits/stdc++.h>
using namespace std;
int ans[1003][1003];
int a;
int num[20009];
int top = 0;
int n;
int main() {
  scanf("%d", &a);
  if (a == 1) {
    printf("2\n");
    printf("NY\nYN\n");
    return 0;
  }
  memset((ans), (0), sizeof((ans)));
  n = 1;
  for (int i = 2; i <= 33; i++) {
    ans[i][i - 1] = ans[i - 1][i] = 1;
  }
  n = 33;
  for (int i = 0; (1 << i) <= a; i++) {
    if (a & (1 << i)) {
      n += 2 * i;
    }
  }
  n++;
  int cur = 34;
  for (int i = 0; (1 << i) <= a; i++) {
    if (a & (1 << i)) {
      if (i == 0)
        ans[33][n] = ans[n][33] = 1;
      else {
        ans[cur][33 - i] = ans[33 - i][cur] = 1;
        ans[cur + 1][33 - i] = ans[33 - i][cur + 1] = 1;
        cur += 2;
        for (int j = 1; j < i; j++) {
          ans[cur][cur - 2] = ans[cur - 2][cur] = 1;
          ans[cur + 1][cur - 1] = ans[cur - 1][cur + 1] = 1;
          ans[cur][cur - 1] = ans[cur - 1][cur] = 1;
          ans[cur + 1][cur - 2] = ans[cur - 2][cur + 1] = 1;
          cur += 2;
        }
        ans[n][cur - 1] = ans[cur - 1][n] = 1;
        ans[n][cur - 2] = ans[cur - 2][n] = 1;
      }
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int a, b;
      a = i, b = j;
      if (a == 2)
        a = n;
      else if (a == n)
        a = 2;
      if (b == 2)
        b = n;
      else if (b == n)
        b = 2;
      printf("%c", ans[a][b] == 0 ? 'N' : 'Y');
    }
    putchar(10);
  }
  return 0;
}
