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 main() {
  int k;
  cin >> k;
  int m = 0, x = 1;
  while (x * 2 <= k) {
    x *= 2;
    m++;
  }
  if (m == 0) {
    cout << 2 << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  int n = (m + 1) * 3;
  cout << n << endl;
  vector<vector<char>> v(n, vector<char>(n, 'N'));
  v[0][2] = 'Y';
  v[0][2 + m] = 'Y';
  v[1 + m][n - 1] = 'Y';
  v[1 + 2 * m][n - 1] = 'Y';
  v[n - 1][1] = 'Y';
  v[n - 2][1] = 'Y';
  for (int i = 2; i < 1 + m; i++) {
    v[i][i + m + 1] = 'Y';
    v[i][i + 1] = 'Y';
    v[i + m][i + 1] = 'Y';
    v[i + m][i + m + 1] = 'Y';
    v[i + 2 * m][i + 2 * m + 1] = 'Y';
  }
  if (x != k) {
    k -= x;
    int j = 0;
    while (j < m) {
      if (((1 << j) & k) != 0) {
        v[j + 2 + m][j + 2 + 2 * m] = 'Y';
      }
      j++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (v[i][j] == 'Y' || v[j][i] == 'Y') {
        v[i][j] = 'Y';
        v[j][i] = 'Y';
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << v[i][j];
    }
    cout << endl;
  }
  return 0;
}