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;
vector<string> G(1000);
int now = 1;
void con(int a, int b) {
  G[a][b] = 'Y';
  G[b][a] = 'Y';
  return;
}
int main() {
  int k;
  cin >> k;
  string sss(1000, 'N');
  for (int i = 0; i < 1000; ++i) G[i] = sss;
  for (int i = 0; k > 0; i++) {
    if (k % 10) {
      for (int p = 1; p <= k % 10; ++p) {
        con(0, now + p);
        con(now + k % 10 + 1, now + p);
      }
      now = now + k % 10 + 1;
      for (int j = 0; j < i; ++j) {
        for (int p = 1; p <= 10; ++p) {
          con(now, now + p);
          con(now + 11, now + p);
        }
        now = now + 11;
      }
      for (int j = 0; j < 20 - 2 * i; ++j) {
        con(now, now + 1);
        now++;
      }
      con(now, 1);
      now++;
    }
    k /= 10;
    if (k == 0) break;
  }
  cout << now << endl;
  for (int i = 0; i < now; ++i) {
    for (int j = 0; j < now; ++j) {
      cout << G[i][j];
    }
    cout << endl;
  }
  return 0;
}