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>
const int N = 1100;
bool mp[N][N];
int n, k;
void addE(int a, int b) { mp[a][b] = mp[b][a] = true; }
void help(int k) {
  while (k--) {
    addE(n, n + 1);
    addE(n, n + 2);
    addE(n + 1, n + 3);
    addE(n + 2, n + 3);
    n += 3;
  }
  addE(n, 2);
}
void build(int rt, int i) {
  if (i < 0) return;
  if (k & (1 << i)) {
    addE(rt, (31 - i) * 3);
  }
  addE(rt, n + 1);
  addE(n + 1, n + 2);
  n += 2;
  build(n, i - 1);
}
int main() {
  n = 3;
  help(30);
  scanf("%d", &k);
  build(1, 30);
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) printf("%c", mp[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}