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 n;
int len;
int a[1010][1010];
void gouzao(int pre1, int pre2, int i, int z) {
  if (i == 0) {
    if (len == 0) {
      a[pre1][2] = a[2][pre1] = 1;
      a[pre2][2] = a[2][pre2] = 1;
    } else {
      if (z == 0)
        a[pre1][2] = a[2][pre1] = 1;
      else {
        int t1 = 1 + 2 * (len - z), t2 = t1 + 1;
        a[pre1][t1] = a[t1][pre1] = 1;
        a[pre1][t2] = a[t2][pre1] = 1;
      }
    }
    return;
  }
  if (len == 0) {
    n++;
    a[pre1][n] = a[n][pre1] = 1;
    a[pre2][n] = a[n][pre2] = 1;
    n++;
    a[pre1][n] = a[n][pre1] = 1;
    a[pre2][n] = a[n][pre2] = 1;
    gouzao(n - 1, n, i - 1, z);
  } else {
    n++;
    a[pre1][n] = a[n][pre1] = 1;
    gouzao(n, n, i - 1, z);
  }
}
int main() {
  int k;
  cin >> k;
  int limit = log2(k);
  n = 2;
  len = 0;
  for (int i = limit; i >= 0; i--) {
    if (k & (1 << i)) {
      gouzao(1, 1, len == 0 ? i : len - i - 1, i);
      if (len == 0) len = limit + 1;
    }
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j] && i != j)
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
}