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;
char adjM[1005][1005];
long long k, n, len, lastOne;
void makeGraph(long long len) {
  for (int(i) = (1); i <= (int)(len - 1); (i)++) {
    adjM[n - 1][n + 1] = adjM[n - 1][n + 2] = adjM[n][n + 1] = adjM[n][n + 2] =
        'Y';
    adjM[n + 1][n - 1] = adjM[n + 2][n - 1] = adjM[n + 1][n] = adjM[n + 2][n] =
        'Y';
    n += 2;
  }
}
void baseCase() {
  long long i = 35;
  while (i > 0 && !(k & (1ll << i))) --i;
  k ^= (1 << i);
  if (i > 0) {
    n = 4;
    adjM[1][3] = adjM[3][1] = adjM[1][4] = adjM[4][1] = 'Y';
    makeGraph(i);
    adjM[2][n] = adjM[2][n - 1] = adjM[n][2] = adjM[n - 1][2] = 'Y';
    len = i, lastOne = n;
    return;
  }
  n = 2;
  adjM[1][2] = adjM[2][1] = 'Y';
}
int main(int argc, const char* argv[]) {
  memset(adjM, 'N', sizeof adjM);
  scanf("%lld", &k);
  baseCase();
  for (long long i = 0; (1ll << i) <= k; ++i) {
    if (k & (1 << i)) {
      ++n, adjM[1][n] = adjM[n][1] = 'Y';
      for (int(j) = (2); j <= (int)(len - i); (j)++)
        ++n, adjM[n - 1][n] = adjM[n][n - 1] = 'Y';
      if (!i)
        adjM[2][n] = adjM[n][2] = 'Y';
      else
        adjM[n][lastOne - 2 * (i - 1)] = adjM[lastOne - 2 * (i - 1)][n] =
            adjM[n][lastOne - 2 * (i - 1) - 1] =
                adjM[lastOne - 2 * (i - 1) - 1][n] = 'Y';
    }
  }
  printf("%lld\n", n);
  for (int(i) = (1); i <= (int)(n); (i)++) {
    for (int(j) = (1); j <= (int)(n); (j)++) printf("%c", adjM[i][j]);
    printf("\n");
  }
  return 0;
}