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 k;
unsigned char ans[1001][1001];
int len[1000], N[1000];
int main() {
  ios_base::sync_with_stdio(0);
  for (int i = 0; i < 1001; ++i)
    for (int j = 0; j < 1001; ++j) ans[i][j] = 'N';
  cin >> k;
  int cur;
  for (cur = 0; 1 << cur < k; ++cur) {
    len[cur] = cur, N[cur] = 1 << (cur - 1);
    ans[cur * 2][cur * 2 + 2] = ans[cur * 2 + 2][cur * 2] = 'Y';
    ans[cur * 2][cur * 2 + 3] = ans[cur * 2 + 3][cur * 2] = 'Y';
    if (cur > 0) {
      ans[cur * 2 + 1][cur * 2 + 2] = ans[cur * 2 + 2][cur * 2 + 1] = 'Y';
      ans[cur * 2 + 1][cur * 2 + 3] = ans[cur * 2 + 3][cur * 2 + 1] = 'Y';
    }
  }
  N[cur] = 1 << (cur - 1);
  ans[cur * 2][1] = ans[1][cur * 2] = 'Y';
  k -= N[cur];
  int LEN = cur + 1;
  int next = LEN * 2;
  if (k == N[cur])
    ans[cur * 2 + 1][1] = ans[1][cur * 2 + 1] = 'Y';
  else {
    for (int c = cur - 1; c > 0; --c)
      if (k >= N[c]) {
        k -= N[c];
        for (int j = 0; j < LEN - len[c]; ++j) {
          if (j == 0) {
            ans[c * 2][next] = ans[next][c * 2] = 'Y';
          } else if (j == LEN - len[c] - 1) {
            ans[next][1] = ans[1][next] = 'Y';
            ++next;
          } else {
            ans[next][next + 1] = ans[next + 1][next] = 'Y';
            ++next;
          }
        }
      }
  }
  cout << (next + 2) << '\n';
  for (int i = 0; i < next + 2; ++i) {
    for (int j = 0; j < next + 2; ++j) cout << ans[i][j];
    cout << '\n';
  }
  return 0;
}