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;
bool ok[1005][1005];
int id[1005];
void add(int a, int b) { ok[a][b] = ok[b][a] = true; }
int main() {
  int k;
  scanf("%d", &k);
  int s = 0;
  for (int j = 1; j <= 30; j++) {
    add(s, j * 3 - 1);
    add(s, j * 3);
    add(j * 3 - 1, j * 3 + 1);
    add(j * 3, j * 3 + 1);
    s = j * 3 + 1;
  }
  s++;
  for (int j = 1; j <= 80; j++) {
    if (j == 1)
      add(1, s);
    else
      add(s - 1, s);
    id[j] = s;
    s++;
  }
  for (int j = 0; j <= 30; j++) {
    if (k >> j & 1) {
      int st = j == 0 ? 0 : j * 3 + 1;
      int end = id[70 - 2 * j];
      add(st, end);
    }
  }
  printf("%d\n", s);
  for (int i = 0; i < s; i++) {
    for (int j = 0; j < s; j++) printf(ok[i][j] ? "Y" : "N");
    puts("");
  }
  return 0;
}