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>
int K;
bool adj[92][92];
void edge(int a, int b) { adj[a - 1][b - 1] = adj[b - 1][a - 1] = true; }
int main() {
  edge(1, 32);
  edge(2, 33);
  edge(2, 63);
  for (int v = 2; v < 31; v++) {
    edge(v + 1, v + 2);
    edge(31 + v, 32 + v);
    edge(31 + v, 62 + v);
    edge(32 + v, 61 + v);
    edge(61 + v, 62 + v);
  }
  scanf("%d", &K);
  if (K & 1) {
    K ^= 1;
    edge(2, 3);
  }
  for (int v = 1; v < 31; v++)
    if ((K >> v) & 1) {
      edge((1 + v) % 31 + 2, v + 32);
      edge((1 + v) % 32 + 2, v + 62);
    }
  printf("92\n");
  for (int i = 0; i < 92; i++) {
    for (int j = 0; j < 92; j++) printf("%c", adj[i][j] ? 'Y' : 'N');
    printf("\n");
  }
  return 0;
}