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 ar[205][205], n, cnt;
void Do(int a, int b) { ar[a][b] = ar[b][a] = 1; }
bool B(int mask, int j) { return mask & (1 << j); }
int main() {
  Do(1, 3);
  Do(1, 4);
  for (int i = 0; i <= 29; i++)
    for (int j = 2 * i + 2; j < 2 * i + 4; j++)
      for (int k = 2 * i + 4; k < 2 * i + 6; k++) Do(j + 1, k + 1);
  for (int i = 66; i < 100; i++) Do(i, i + 1);
  Do(100, 2);
  cin >> n;
  for (int i = 0; i < 30; i++) {
    if (B(n, i)) Do((i + 1) * 2 + 2, 67 + i + 1);
  }
  cout << 100 << endl;
  for (int i = 1; i <= 100; i++) {
    for (int j = 1; j <= 100; j++) {
      if (ar[i][j] == 1 && i != j)
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
}