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.
#include <bits/stdc++.h>
using namespace std;
long long int n, i, lev, cnt, j, k, l, maxi;
long long int gr[5000][5000];
vector<long long int> v[100];
int main() {
  cin >> n;
  i = 3;
  while (1) {
    v[0].push_back(1);
    lev = 1;
    cnt = 2;
    for (;;) {
      if (cnt > n) break;
      v[lev].push_back(i);
      i++;
      v[lev].push_back(i);
      i++;
      lev++;
      cnt *= 2;
    }
    maxi = max(maxi, lev);
    for (j = lev; j < maxi; j++) {
      v[j].push_back(i);
      i++;
    }
    n -= cnt / 2;
    v[j].push_back(2);
    for (j = 0; j < maxi; j++) {
      for (k = 0; k < v[j].size(); k++) {
        for (l = 0; l < v[j + 1].size(); l++) {
          gr[v[j][k]][v[j + 1][l]] = 1;
          gr[v[j + 1][l]][v[j][k]] = 1;
        }
      }
    }
    cnt = cnt / 4;
    lev = 1;
    while (n) {
      if (cnt & n) {
        long long int x = v[lev - 1].size() - 1;
        gr[v[lev - 1][x]][i] = 1;
        gr[i][v[lev - 1][x]] = 1;
        for (j = 0; j < v[lev + 1].size(); j++) {
          gr[i][v[lev + 1][j]] = 1;
          gr[v[lev + 1][j]][i] = 1;
        }
        v[lev].push_back(i);
        i++;
        n -= cnt;
      } else if (n) {
        v[lev].push_back(i);
        long long int x = v[lev - 1].size() - 1;
        gr[v[lev - 1][x]][i] = 1;
        gr[i][v[lev - 1][x]] = 1;
        i++;
      }
      lev++;
      cnt /= 2;
    }
    break;
  }
  printf("%lld\n", i - 1);
  for (j = 1; j < i; j++) {
    for (k = 1; k < i; k++) {
      if (gr[j][k])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}
