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 main() {
  int n, m, k;
  cin >> k;
  std::vector<bool> e(1001 * 1001, false);
  n = 2;
  int d = 0;
  while (k > 0) {
    int s = k % 10;
    if (k == 10) {
      s = 10;
      k = 0;
    }
    k /= 10;
    if (s == 0) {
      d++;
      continue;
    }
    int newn = n;
    for (int i = 1; i <= s; i++) {
      e[1 * 1001 + n + i] = true;
      e[(n + i) * 1001 + 1] = true;
    }
    newn += s;
    for (int k = 1; k <= d; k++) {
      for (int i = 0; i < s; i++)
        for (int j = 1; j <= 10; j++) {
          e[(newn + j) * 1001 + newn - i] = true;
          e[(newn - i) * 1001 + newn + j] = true;
        }
      newn += 10;
      s = 10;
    }
    for (int k = 1; k <= 8 - d; k++) {
      for (int i = 0; i < s; i++)
        for (int j = 1; j <= 1; j++) {
          e[(newn + j) * 1001 + newn - i] = true;
          e[(newn - i) * 1001 + newn + j] = true;
        }
      newn += 1;
      s = 1;
    }
    for (int i = 0; i < s; i++) {
      e[2 * 1001 + newn - i] = true;
      e[(newn - i) * 1001 + 2] = true;
    }
    n = newn;
    d++;
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (e[i * 1001 + j])
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
  return 0;
}
