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 lens;
int nums;
int st;
void chu(int x) {
  lens = 0;
  nums = 0;
  while (x) {
    if (x % 3) {
      biao[lens] = x % 3;
      nums++;
    } else
      biao[lens] = 0;
    lens++;
    x = x / 3;
  }
}
void dos() {
  int i, j;
  for (i = 1; i <= st; i++) {
    for (j = 1; j <= st; j++) {
      if (maps[i][j]) maps[j][i] = 1;
    }
  }
}
int main() {
  int n, i, j, k;
  int ans;
  while (~scanf("%d", &n)) {
    chu(n);
    st = 3;
    vector<int> vec;
    for (i = 0; i < lens; i++) {
      if (biao[i]) {
        int tt;
        tt = 0;
        vec.clear();
        if (biao[i] == 1) {
          maps[1][st] = 1;
          vec.push_back(st);
          st++;
        } else {
          maps[1][st] = 1;
          vec.push_back(st);
          st++;
          maps[1][st] = 1;
          vec.push_back(st);
          st++;
        }
        for (j = 0; j <= i; j++) {
          for (k = 0; k < vec.size(); k++) {
            maps[vec[k]][st] = 1;
            maps[vec[k]][st + 1] = 1;
            maps[vec[k]][st + 2] = 1;
          }
          vec.clear();
          vec.push_back(st);
          vec.push_back(st + 1);
          vec.push_back(st + 2);
          st = st + 3;
        }
        st = st - 3;
        for (k = 1; k <= 1000; k++) {
          maps[k][st + 1] = 0;
          maps[k][st + 2] = 0;
        }
        for (; j < lens; j++) {
          maps[st][st + 1] = 1;
          st++;
        }
        maps[st][2] = 1;
        st++;
      }
    }
    st--;
    dos();
    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;
}