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>
bool ans[1005][1005];
bool in[55];
int n, m;
void make_road(int v1, int v2) { ans[v1][v2] = ans[v2][v1] = true; }
int main() {
  int k, length = 1;
  scanf("%d", &k);
  while (k != 0) {
    in[m++] = k % 2;
    k /= 2;
  }
  int start = n++, end = n++;
  make_road(start, end);
  for (int i = m - 2; i >= 0; i--) {
    int old_start = n++;
    for (int j = 0; j < n; j++) {
      if (ans[start][j]) make_road(old_start, j);
      ans[start][j] = ans[j][start] = false;
    }
    make_road(start, n), n++;
    make_road(old_start, n - 1);
    make_road(start, n), n++;
    make_road(old_start, n - 1);
    length += 2;
    if (in[i]) {
      make_road(start, n), n++;
      for (int j = 0; j < length - 2; j++) make_road(n - 1, n), n++;
      make_road(n - 1, end);
    }
  }
  printf("%d\n", n);
  for (int i = 0; i < n; i++, printf("\n"))
    for (int j = 0; j < n; j++) printf("%c", (ans[i][j]) ? 'Y' : 'N');
  return 0;
}