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 maps[1001][1001];
int biao[101];
int cal(int x) {
  int len = 0;
  while (x > 0) {
    biao[len++] = x % 3;
    x /= 3;
  }
  return len;
}
int main() {
  int n, i, j, k;
  int ans;
  scanf("%d", &n);
  int st = 3;
  vector<int> vec;
  int lens = cal(n);
  for (i = 0; i < lens; i++) {
    if (biao[i] > 0) {
      vec.clear();
      if (biao[i] == 1) {
        maps[1][st] = 1;
        vec.push_back(st++);
      } else {
        maps[1][st] = 1;
        vec.push_back(st++);
        maps[1][st] = 1;
        vec.push_back(st++);
      }
      for (int j = 0; j <= i; ++j) {
        for (int k = 0; k < vec.size(); ++k) {
          maps[vec[k]][st] = maps[vec[k]][st + 1] = maps[vec[k]][st + 2] = 1;
        }
        vec.clear();
        vec.push_back(st), vec.push_back(st + 1), vec.push_back(st + 2);
        st += 3;
      }
      st -= 3;
      for (int j = 1; j <= 1000; ++j) {
        maps[j][st + 1] = maps[j][st + 2] = 0;
      }
      int j = i + 1;
      while (j < lens) {
        ++j;
        maps[st][st + 1] = 1;
        st++;
      }
      maps[st++][2] = 1;
    }
  }
  st--;
  for (i = 1; i <= st; i++) {
    for (j = 1; j <= st; j++) {
      if (maps[i][j]) maps[j][i] = 1;
    }
  }
  cout << st << endl;
  for (i = 1; i <= st; i++) {
    for (j = 1; j <= st; j++) {
      if (maps[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}