Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m;
int a[1010][1010];
inline void Do(int x, int y) { a[x][y] = a[y][x] = 1; }
inline void work(int x) {
  int i;
  if (!x) {
    Do(1, 3);
    return;
  }
  Do(1, n + 1), Do(1, n + 2);
  for (i = 1; i <= x - 1; i++) {
    Do(n + 1, n + 3), Do(n + 1, n + 4);
    Do(n + 2, n + 3), Do(n + 2, n + 4);
    n += 2;
  }
  int A = 3 + x;
  Do(n + 1, A), Do(n + 2, A);
  n += 2;
}
int main() {
  scanf("%d", &m);
  n = 2;
  for (i = 1; i <= 40; i++) Do(n + 1, n + 2), n++;
  Do(n + 1, 2), n++;
  for (i = 0; m; m >>= 1, i++)
    if (m & 1) work(i);
  printf("%d\n", n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) printf("%c", a[i][j] ? 'Y' : 'N');
    printf("\n");
  }
  return 0;
}