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.
#include <bits/stdc++.h>
using namespace std;
const int N = 31;
int node[N][3], idx[N][3], cnt = 2;
bool in[101][101];
int main() {
  node[0][0] = node[0][1] = ++cnt;
  node[0][2] = ++cnt;
  for (int i = 1; i < N; i++) {
    node[i][0] = ++cnt;
    node[i][1] = ++cnt;
    node[i][2] = ++cnt;
    in[node[i][0]][node[i - 1][0]] = 1;
    in[node[i][0]][node[i - 1][1]] = 1;
    in[node[i - 1][0]][node[i][0]] = 1;
    in[node[i - 1][1]][node[i][0]] = 1;
    in[node[i][1]][node[i - 1][0]] = 1;
    in[node[i][1]][node[i - 1][1]] = 1;
    in[node[i - 1][0]][node[i][1]] = 1;
    in[node[i - 1][1]][node[i][1]] = 1;
    in[node[i][2]][node[i - 1][2]] = 1;
    in[node[i - 1][2]][node[i][2]] = 1;
  }
  in[node[0][0]][2] = 1;
  in[2][node[0][0]] = 1;
  in[1][node[N - 1][2]] = 1;
  in[node[N - 1][2]][1] = 1;
  int k;
  cin >> k;
  for (int i = 0; i < N; i++)
    if (k & (1 << i)) {
      in[node[i][0]][node[i][2]] = 1;
      in[node[i][2]][node[i][0]] = 1;
      in[node[i][1]][node[i][2]] = 1;
      in[node[i][2]][node[i][1]] = 1;
    }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++, cout << endl)
    for (int j = 1; j <= cnt; j++) cout << ((in[i][j]) ? 'Y' : 'N');
  return 0;
}
