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 k, sum, Sum[2005], ans;
int a[2005][2005], la1, la2, now1, now2, m;
int main() {
  scanf("%lld", &k);
  sum = 1;
  ans = 1;
  la1 = la2 = 2;
  now1 = 3;
  now2 = 4;
  for (int i = 1; i <= 40; i++) {
    m = max(m, now2);
    a[la1][now1] = a[la1][now2] = a[la2][now1] = a[la2][now2] = 1;
    a[now1][la1] = a[now2][la1] = a[now1][la2] = a[now2][la2] = 1;
    Sum[now1] = Sum[now2] = sum;
    la1 = now1;
    la2 = now2;
    now1 += 2;
    now2 += 2;
    sum *= 2;
    if (sum > k) break;
  }
  int la = 1;
  bool fl = false;
  while (k) {
    if (sum > k) {
      if (fl) {
        m++;
        a[la][m] = a[m][la] = 1;
        la = m;
      }
      sum >>= 1;
      continue;
    }
    fl = true;
    m++;
    a[la][m] = a[m][la] = 1;
    la = m;
    int i;
    for (i = 1; i <= m; i++)
      if (Sum[i] == sum) break;
    a[m][i] = a[i][m] = 1;
    k -= sum;
    sum >>= 1;
  }
  printf("%d\n", m);
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) putchar(a[i][j] ? 'Y' : 'N');
    puts("");
  }
}
