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>
using namespace std;
int k, n, g[1000][1000];
void cal(int k) {
  while (k) {
    k >>= 1;
    n++;
  }
  n--;
}
int main() {
  int i, j, t = 1, x = 3, p = 3;
  cin >> k;
  if (k == 1) {
    puts("2");
    puts("NY");
    puts("YN");
    return 0;
  }
  cal(k);
  for (i = 0; i < n; i++) {
    g[t][x] = g[t][x + 1] = 1;
    t = x + 2;
    g[t][x] = g[t][x + 1] = 1;
    x = x + 3;
  }
  g[t][2] = 1;
  for (i = 0; i < n; i++)
    if (k & 1 << i) {
      g[1][x] = 1;
      for (j = 0; j < n - i - 1; j++) {
        g[x][x + 1] = 1;
        g[x + 1][x + 2] = 1;
        x = x + 2;
      }
      g[x][(n - i) * 3 + 2] = 1;
      x++;
    }
  cout << x << endl;
  for (i = 1; i <= x; i++, cout << endl)
    for (j = 1; j <= x; j++)
      if (g[i][j] || g[j][i])
        printf("Y");
      else
        printf("N");
  return 0;
}