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 T = 1e3 + 5;
int g[T][T];
void add(int a, int b) { g[a][b] = g[b][a] = 1; }
int last, K, a, b, i, j, N;
int main() {
  cin >> K;
  add(0, 2);
  add(0, 3);
  for (i = 2; i <= 30; i++) {
    add(2 * i, 2 * i - 1);
    add(2 * i, 2 * i - 2);
    add(2 * i + 1, 2 * i - 1);
    add(2 * i + 1, 2 * i - 2);
  }
  N = 2 * 30 + 1;
  for (i = 30; i >= 0; i--)
    if (K & (1 << (i - 1))) {
      last = 2 * i;
      for (j = i; j < 30; j++) {
        N++;
        add(N, last);
        last = N;
      }
      add(last, 1);
    }
  cout << N + 1 << endl;
  for (i = 0; i <= N; i++) {
    for (j = 0; j <= N; j++)
      if (g[i][j])
        printf("%c", 'Y');
      else
        printf("%c", 'N');
    printf("\n");
  }
}
