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;
struct __timestamper {};
char s[1010][1010];
int n;
void addedge(int u, int v) {
  assert(u != v);
  assert(s[u][v] != 'Y');
  s[u][v] = s[v][u] = 'Y';
}
int solve(int f, int t, int k) {
  for (int i = 2; i * i <= k; i++)
    if (k % i == 0) {
      int ne = n++;
      assert(n <= 1000);
      int tmp = solve(f, ne, i);
      tmp += solve(ne, t, k / i);
      return tmp;
    }
  if (k == 1) {
    int ne = n++;
    addedge(f, ne);
    addedge(ne, t);
    return 2;
  }
  int tmp = solve(f, t, k - 1);
  for (int i = 0; i < tmp; i++) {
    int nf = (i == tmp - 1 ? t : n++);
    addedge(f, nf);
    f = nf;
  }
  return tmp;
}
int main() {
  int k;
  scanf("%d", &k);
  n = 2;
  memset(s, 'N', sizeof(s));
  solve(0, 1, k);
  printf("%d\n", n);
  for (int i = 0; i < n; i++) s[i][n] = 0, puts(s[i]);
  return 0;
}