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, lg;
bool g[1010][1010];
void add(int a, int b) {
  g[a][b] = true;
  g[b][a] = true;
}
int main() {
  cin >> k;
  for (lg = 0; k >> (lg + 1); lg++)
    ;
  if (k > 1) {
    add(1, 3);
    add(1, 4);
    add(2 * lg + 1, 2);
    add(2 * lg + 2, 2);
    add(99 + lg, 2);
  } else
    add(1, 2);
  if (k & 1) add(1, 100);
  for (int i = 0; i < lg - 1; i++) {
    int x = 3 + 2 * i;
    add(x, x + 2);
    add(x, x + 3);
    add(x + 1, x + 2);
    add(x + 1, x + 3);
    add(100 + i, 101 + i);
    if ((k >> (i + 1)) & 1) {
      add(x, 101 + i);
      add(x + 1, 101 + i);
    }
  }
  cout << 200 << '\n';
  for (int i = 1; i <= 200; i++) {
    for (int j = 1; j <= 200; j++) cout << (g[i][j] ? 'Y' : 'N');
    cout << '\n';
  }
}