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;
int powInt(int x, int y) {
  int r = 1;
  while (y > 0) {
    if (y & 1) r = (r * x);
    x = (x * x);
    y /= 2;
  }
  return r;
}
int qrt(long k, int ex) {
  int res = 1;
  while (powInt(res, ex) <= k) {
    res++;
  }
  return res - 1;
}
int main() {
  int k;
  cin >> k;
  vector<int> nds;
  int ex = 3;
  int sum, n;
  do {
    int tmpk = k;
    sum = 0;
    while (tmpk > 0) {
      int tmp = qrt((long)tmpk, ex);
      nds.push_back(tmp);
      tmpk -= powInt(tmp, ex);
    }
    for (int((i)) = (0); ((i)) < ((nds.size())); ((i))++) {
      sum += nds[i];
    }
    n = ex * sum + 2;
    if (n > 1000) {
      nds.clear();
      ex++;
    }
  } while (n > 1000);
  string init(n, 'N');
  vector<string> g(n, init);
  for (int(i) = (2); (i) < (sum + 2); (i)++) {
    g[0][i] = 'Y';
    g[i][0] = 'Y';
    g[1][i + (ex - 1) * sum] = 'Y';
    g[i + (ex - 1) * sum][1] = 'Y';
  }
  int offset = 2;
  cout << n << endl;
  for (int((i)) = (0); ((i)) < ((nds.size())); ((i))++) {
    int c = nds[i];
    for (int(j) = (offset); (j) < (offset + c); (j)++) {
      for (int(q) = (0); (q) < (ex - 1); (q)++) {
        for (int(p) = (0); (p) < (nds[i]); (p)++) {
          g[j + q * sum][(q + 1) * sum + p + offset] = 'Y';
          g[(q + 1) * sum + p + offset][j + q * sum] = 'Y';
        }
      }
    }
    offset += c;
  }
  for (int((i)) = (0); ((i)) < ((n)); ((i))++) {
    cout << g[i] << endl;
  }
  return 0;
}
