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 = 200;
const int M = 30;
bool g[N + 10][N + 10];
void solve() {
  int target;
  cin >> target;
  memset(g, 0, sizeof(g));
  if (target & 1) g[1][3] = 1;
  g[M * 3][2] = 1;
  g[1][4] = g[1][5] = 1;
  for (int i = 1; i < M; i++) {
    int a = i * 3, b = a + 3;
    g[a][b] = 1;
    for (int j = 1; j <= 2; j++) {
      for (int k = 1; k <= 2; k++) {
        g[a + j][b + k] = 1;
      }
    }
  }
  target >>= 1;
  int layer = 1;
  while (target) {
    if (target & 1) {
      int a = layer * 3, b = a + 3;
      g[a + 1][b] = g[a + 2][b] = 1;
    }
    target >>= 1;
    layer++;
  }
  cout << N << endl;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      if (g[i][j] || g[j][i])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
}
int main() {
  solve();
  return 0;
}