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 cur, num, msk;
bool adj[250][250];
void dfs(int nd, int dep, int sep) {
  if (dep > 30) return;
  adj[nd][cur] = true;
  adj[cur][nd] = true;
  adj[nd][cur + 1] = true;
  adj[cur + 1][nd] = true;
  adj[cur][cur + 2] = true;
  adj[cur + 2][cur] = true;
  adj[cur + 1][cur + 2] = true;
  adj[cur + 2][cur + 1] = true;
  if (msk & (1 << dep)) {
    adj[nd][sep] = true;
    adj[sep][nd] = true;
  }
  nd = cur + 2;
  cur += 3;
  dfs(nd, dep + 1, sep + 2);
}
int main() {
  scanf("%d", &num);
  cur = 3;
  for (int i = 0; i < (int)100; i++) {
    adj[cur][cur + 1] = true;
    adj[cur + 1][cur] = true;
    cur++;
  }
  adj[cur][2] = true;
  adj[2][cur] = true;
  cur++;
  for (int i = 30; i >= 0; i--) {
    if (num < (1 << i)) continue;
    num -= (1 << i);
    msk |= (1 << i);
  }
  dfs(1, 0, 3);
  printf("249\n");
  for (int i = 1; i < (int)250; i++) {
    for (int j = 1; j < (int)250; j++) printf("%c", adj[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}