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.
#include <bits/stdc++.h>
using namespace std;
long long n = 2, k, len = 29, a[2111][2111];
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  cin >> k;
  long long x = 1, y = 0, z = 0, calc = 1;
  for (long long i = 1; i <= len; i++) {
    if (calc <= k) calc *= 3;
    if (calc <= k) {
      a[x][n + 1] = a[n + 1][x] = 1;
      a[x][n + 2] = a[n + 2][x] = 1;
      a[x][n + 3] = a[n + 3][x] = 1;
      a[y][n + 1] = a[n + 1][y] = 1;
      a[y][n + 2] = a[n + 2][y] = 1;
      a[y][n + 3] = a[n + 3][y] = 1;
      a[z][n + 1] = a[n + 1][z] = 1;
      a[z][n + 2] = a[n + 2][z] = 1;
      a[z][n + 3] = a[n + 3][z] = 1;
      x = n + 1, y = n + 2, z = n + 3;
      n += 3;
    } else {
      a[x][n + 1] = a[n + 1][x] = 1;
      a[y][n + 1] = a[n + 1][y] = 1;
      a[z][n + 1] = a[n + 1][z] = 1;
      x = n + 1, y = z = 0;
      n++;
    }
  }
  a[x][2] = a[2][x] = 1;
  a[y][2] = a[2][y] = 1;
  a[z][2] = a[2][z] = 1;
  k -= calc / 3;
  for (long long i = len; i >= 0; i--) {
    if ((k >> i) & 1) {
      long long l1 = 1, l2 = 0;
      for (long long j = 1; j <= len; j++) {
        if (j <= i) {
          a[l1][n + 1] = a[n + 1][l1] = a[l1][n + 2] = a[n + 2][l1] = 1;
          a[l2][n + 1] = a[n + 1][l2] = a[l2][n + 2] = a[n + 2][l2] = 1;
          l1 = n + 1;
          l2 = n + 2;
          n += 2;
        } else {
          a[l1][n + 1] = a[n + 1][l1] = 1;
          a[l2][n + 1] = a[n + 1][l2] = 1;
          l1 = n + 1;
          l2 = 0;
          n++;
        }
      }
      a[l1][2] = a[2][l1] = 1;
      a[l2][2] = a[2][l2] = 1;
    }
  }
  cout << n;
  cout << endl;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) cout << (a[i][j] == 1 ? "Y" : "N");
    cout << endl;
  }
}
