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;
bool A[1001][1001] = {{false}};
long long h[32], g[32], f[32], c = 3;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long pv[2] = {1, 1};
  f[0] = c++;
  g[0] = 1;
  long long k;
  cin >> k;
  long long n = 1;
  for (long long x = 1; x < k; x *= 2) n++;
  for (long long i = 1; i < n; i++) {
    long long a = c++;
    long long b = c++;
    g[i] = b;
    h[i] = a;
    A[pv[0]][a] = A[a][pv[0]] = A[pv[1]][a] = A[a][pv[1]] = true;
    A[pv[0]][b] = A[b][pv[0]] = A[pv[1]][b] = A[b][pv[1]] = true;
    f[i] = c++;
    A[f[i]][f[i - 1]] = A[f[i - 1]][f[i]] = true;
    pv[0] = a;
    pv[1] = b;
  }
  A[2][f[n - 1]] = A[f[n - 1]][2] = true;
  for (long long b = 0; b < n; b++) {
    if (k & (1 << b)) {
      A[g[b]][f[b]] = A[f[b]][g[b]] = true;
      A[h[b]][f[b]] = A[f[b]][h[b]] = true;
    }
  }
  cout << c - 1 << '\n';
  for (long long i = 1; i < c; i++) {
    for (long long j = 1; j < c; j++) {
      cout << (A[i][j] ? 'Y' : 'N');
    }
    cout << '\n';
  }
}
