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 k, n, g[3005][3005];
void ed(int a, int b) { g[a][b] = g[b][a] = 1; }
int main() {
  scanf("%d", &k);
  n = 3;
  int pl = 0, ps = 2;
  for (int i = 30; i >= 0; --i) {
    int a1 = n++;
    int a2 = n++;
    int t = n++;
    ed(t, a1);
    ed(t, a2);
    ed(a1, ps);
    ed(a2, ps);
    if (pl) pl += 2;
    if (k & (1 << i)) {
      if (!pl) pl = 1;
      int pr = t;
      for (int j = 0; j < pl; ++j) {
        int v = n++;
        ed(pr, v);
        pr = v;
      }
      ed(pr, 1);
    }
    ps = t;
  }
  ed(0, ps);
  printf("%d\n", n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) printf("%c", g[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}