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 maxn = 1005;
int k, n, ak[100], ak0;
bool mp[maxn][maxn];
int p[maxn];
int main() {
  scanf("%d", &k);
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  for (; k; k >>= 1) ak[++ak0] = (k & 1);
  n = 1;
  int now = 1, len = 0;
  for (int i = ak0 - 1; i >= 1; i--) {
    len += 2;
    mp[n + 1][now] = mp[now][n + 1] = 1;
    mp[n + 2][now] = mp[now][n + 2] = 1;
    n += 3;
    mp[n - 2][n] = mp[n][n - 2] = 1;
    mp[n - 1][n] = mp[n][n - 1] = 1;
    now = n;
    if (ak[i]) {
      int last = 1;
      for (int j = 1; j <= len - 1; j++) {
        n++;
        mp[last][n] = mp[n][last] = 1;
        last = n;
      }
      mp[last][now] = mp[now][last] = 1;
    }
  }
  for (int i = 1; i <= n; i++) p[i] = i;
  p[2] = now, p[now] = 2;
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) putchar(mp[p[i]][p[j]] ? 'Y' : 'N');
    putchar('\n');
  }
}