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;
char a[151][151];
int k, n = 150, i, j;
void make(int x, int y) {
  a[x][y] = 'Y';
  a[y][x] = 'Y';
}
int main() {
  cin >> k;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) a[i][j] = 'N';
  make(1, 3);
  for (i = 3; i < 70; ++i) {
    if (i & 1)
      make(i, i + 2), make(i, i + 1);
    else
      make(i, i + 3), make(i, i + 2);
  }
  for (i = 100; i < 150; ++i) make(i, i + 1);
  make(i, 2);
  for (j = 0; j < 32; ++j)
    if ((1 << j) & k) make((j + 2) * 2, 100 + j);
  cout << n << "\n";
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) cout << a[i][j];
    cout << "\n";
  }
  return 0;
}