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;
char grid[1010][1010];
long long ptr[1010];
void add(long long u, long long v) { grid[u][v] = grid[v][u] = 'Y'; }
int main() {
  long long N;
  scanf("%lld", &N);
  for (long long i = 1; i <= 1000; i++)
    for (long long j = 1; j <= 1000; j++) grid[i][j] = 'N';
  long long mx = 0, v = 2;
  for (long long i = 35; i >= 0; i--) {
    if ((N & (1LL << i))) {
      mx = max(mx, i);
      v++;
      add(1, v);
      for (long long j = 1; j <= i; j++) {
        if (i != mx) break;
        ptr[mx - j + 1] = v;
        add(v, v + 1);
        add(v + 1, v + 3);
        add(v, v + 2);
        add(v + 2, v + 3);
        v += 3;
        ptr[mx - j] = v;
      }
      for (long long j = 1; j < 2 * mx - 2 * i; j++) {
        add(v, v + 1);
        v++;
      }
      if (i == mx)
        add(v, 2);
      else
        add(v, ptr[i]);
    }
  }
  cout << v << endl;
  for (long long i = 1; i <= v; i++) {
    for (long long j = 1; j <= v; j++) printf("%c", grid[i][j]);
    printf("\n");
  }
  return 0;
}
