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;
const int maxn = 1010;
long long a[maxn][maxn], sl[maxn], b[maxn][maxn];
queue<long long> q;
void pov(long long x, long long y) {
  a[x][y] = 1;
  a[y][x] = 1;
}
int main() {
  long long k, x = 0, n = 1;
  scanf("%I64d", &k);
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  while (1 << x <= k) x++;
  x--;
  for (long long i = 1; i <= x + 1; i++) {
    n += 2;
    pov(n - 3, n - 1);
    pov(n - 3, n);
    pov(n - 2, n - 1);
    pov(n - 2, n);
  }
  for (long long i = x + 1; i; i--)
    if (1 & (k >> (i - 1))) {
      long long nc = i << 1;
      n++;
      pov(n, nc);
      for (long long j = i; j <= x; j++) {
        n++;
        pov(n, n - 1);
      }
      q.push(n);
    }
  n++;
  while (!q.empty()) {
    long long c = q.front();
    pov(n, c);
    q.pop();
  }
  for (long long i = 1; i <= n; i++) sl[i] = i;
  sl[2] = n;
  sl[n] = 2;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++) b[sl[i]][sl[j]] = a[i][j];
  printf("%I64d\n", n);
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++)
      if (b[i][j])
        printf("Y");
      else
        printf("N");
    printf("\n");
  }
  return 0;
}
