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;
long long a[1010][1010], b[1010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, cnt = 3, k, pre = 1, cur, tp, l = 1;
  cin >> k;
  if (k == 1) {
    cout << "2\nNY\nYN\n";
    return 0;
  }
  tp = k;
  while (tp > 1) {
    b[l++] = tp % 2;
    tp /= 2;
  }
  l--;
  for (i = 1; i <= l; i++) {
    tp = pre;
    if (b[i] == 1) {
      for (j = i; j < l; j++) {
        cur = cnt + 1;
        a[pre][cnt] = 1;
        a[cnt][pre] = 1;
        a[cnt][cur] = 1;
        a[cur][cnt++] = 1;
        pre = cnt++;
      }
      a[pre][cnt] = 1;
      a[cnt][pre] = 1;
      a[cnt][2] = 1;
      a[2][cnt++] = 1;
    }
    pre = tp;
    if (i != l)
      cur = cnt + 2;
    else
      cur = 2;
    a[pre][cnt] = 1;
    a[cnt][pre] = 1;
    a[cur][cnt] = 1;
    a[cnt++][cur] = 1;
    a[pre][cnt] = 1;
    a[cnt][pre] = 1;
    a[cur][cnt] = 1;
    a[cnt++][cur] = 1;
    pre = cnt++;
  }
  cnt--;
  cnt--;
  cout << cnt << "\n";
  for (i = 1; i <= cnt; i++) {
    for (j = 1; j <= cnt; j++)
      if (a[i][j] == 1)
        cout << "Y";
      else
        cout << "N";
    cout << "\n";
  }
  return 0;
}
