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 e[1000][1000];
void add(int a, int b) {
  e[a][b] = 1;
  e[b][a] = 1;
}
int main() {
  int i, j, k;
  scanf("%d", &k);
  for (i = 0; i < 1000; i++)
    for (j = 0; j < 1000; j++) e[i][j] = 0;
  add(0, 2);
  for (i = 0; i < 30; i++) {
    add(i * 3 + 2, i * 3 + 3);
    add(i * 3 + 2, i * 3 + 4);
    add(i * 3 + 3, i * 3 + 5);
    add(i * 3 + 4, i * 3 + 5);
  }
  int n = 30 * 3 + 3;
  for (i = 0; i < 30 * 2 - 1; i++) {
    add(n + i, n + i + 1);
  }
  add(n + 30 * 2 - 1, 1);
  int c = 0;
  while (k > 0) {
    if (k % 2 == 1) add(c * 3 + 2, n + c * 2);
    c++;
    k /= 2;
  }
  n += 30 * 2;
  printf("%d\n", n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) printf("%c", e[i][j] == 1 ? 'Y' : 'N');
    puts("");
  }
  return 0;
}