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 n = 2, k, last_cnt = 0, cnt = 4, level = 6, last_level = 2,
      point_level = -1;
  cin >> k;
  vector<vector<bool>> a(100, vector<bool>(100, false));
  for (int i = int(2); i < int(6); i++) {
    a[i][0] = true;
    a[0][i] = true;
    last_cnt++;
    if (last_cnt == k) {
      for (int j = 2; j <= i; j++) {
        a[j][1] = true;
        a[1][j] = true;
      }
      cout << i + 1 << endl;
      for (int r = 0; r < int(i + 1); r++) {
        for (int t = 0; t < int(i + 1); t++) cout << (a[r][t] ? "Y" : "N");
        cout << endl;
      }
      return 0;
    }
  }
  while (true) {
    if (last_cnt + cnt <= k || (level - 2) % 4 == 0) {
      if ((level - 2) % 4 != 0) last_cnt += cnt;
      for (int i = last_level; i < last_level + 4; i++) {
        a[level][i] = true;
        a[i][level] = true;
      }
      level++;
    } else {
      point_level = level;
      level--;
      while ((level - 1) % 4 != 0) {
        a[level][1] = true;
        a[1][level] = true;
        level--;
      }
      a[point_level][1] = true;
      a[1][point_level] = true;
      level = point_level;
      cnt /= 4;
      while (true) {
        if (last_cnt == k) break;
        for (int i = last_level; i < last_level + 4; i++) {
          if (last_cnt + cnt <= k) {
            a[level][i] = true;
            a[i][level] = true;
            last_cnt += cnt;
          } else
            break;
        }
        cnt /= 4;
        last_level -= 4;
        a[level][level + 1] = true;
        a[level + 1][level] = true;
        level++;
      }
      break;
    }
    if ((level - 2) % 4 == 0) {
      cnt *= 4;
      last_level += 4;
    }
  }
  cout << level << endl;
  for (int r = 0; r < int(level); r++) {
    for (int t = 0; t < int(level); t++) cout << (a[r][t] ? "Y" : "N");
    cout << endl;
  }
}