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 main() {
  char adj[(3 * (32) + 3)][(3 * (32) + 3)];
  for (int i = 0; i < (3 * (32) + 3); i++)
    for (int j = 0; j < (3 * (32) + 3); j++) adj[i][j] = 'N';
  (adj[1][3] = 'Y', adj[3][1] = 'Y');
  (adj[1][4] = 'Y', adj[4][1] = 'Y');
  (adj[(3 * (32) + 3) - 1][2] = 'Y', adj[2][(3 * (32) + 3) - 1] = 'Y');
  for (int i = 0; i < (32) - 1; i++) {
    (adj[3 + i * 3][6 + i * 3] = 'Y', adj[6 + i * 3][3 + i * 3] = 'Y');
    (adj[3 + i * 3][7 + i * 3] = 'Y', adj[7 + i * 3][3 + i * 3] = 'Y');
    (adj[4 + i * 3][6 + i * 3] = 'Y', adj[6 + i * 3][4 + i * 3] = 'Y');
    (adj[4 + i * 3][7 + i * 3] = 'Y', adj[7 + i * 3][4 + i * 3] = 'Y');
    (adj[5 + i * 3][8 + i * 3] = 'Y', adj[8 + i * 3][5 + i * 3] = 'Y');
  }
  int k;
  cin >> k;
  for (int idx = 4; k != 0; k = k >> 1, idx = idx + 3) {
    if (k & 1) {
      (adj[idx][idx + 1] = 'Y', adj[idx + 1][idx] = 'Y');
    }
  }
  printf("%d\n", (3 * (32) + 3) - 1);
  for (int i = 1; i < (3 * (32) + 3); i++) {
    for (int j = 1; j < (3 * (32) + 3); j++) {
      putchar(adj[i][j]);
    }
    putchar('\n');
  }
  return 0;
}