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 bit[33];
int map[1111][1111];
int pr[33], to[33];
int main() {
  int k;
  scanf("%d", &k);
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  for (int i = 0; i <= 30; i++)
    if ((k >> i) & 1) bit[i] = true;
  bool fst = true;
  int p = 3;
  for (int i = 30; i >= 0; i--) {
    if (!bit[i]) continue;
    if (fst) {
      fst = false;
      int l, r;
      l = p++;
      r = p++;
      map[1][l] = true;
      map[1][r] = true;
      pr[1] = l;
      for (int j = 1; j != i; j++) {
        map[l][p] = true;
        map[l][p + 1] = true;
        map[r][p] = true;
        map[r][p + 1] = true;
        l = p++;
        r = p++;
        pr[j + 1] = l;
      }
      map[l][2] = true;
      map[r][2] = true;
      to[0] = p++;
      for (int j = 1; j != i; j++) {
        to[j] = p++;
        map[to[j - 1]][to[j]] = true;
      }
      map[to[i - 1]][2] = true;
    } else {
      if (i == 0) {
        map[1][to[0]] = true;
      } else {
        int fm = pr[i];
        map[fm][to[i]] = true;
        map[fm + 1][to[i]] = true;
      }
    }
  }
  printf("%d\n", p - 1);
  for (int i = 1; i != p; i++) {
    for (int j = 1; j != p; j++) {
      if (map[i][j] || map[j][i])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
  return 0;
}