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.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1005;
bool ans[MAX][MAX];
void add(int u, int v) { ans[u][v] = ans[v][u] = true; }
int main() {
  ios::sync_with_stdio(false);
  int k, p2[35];
  cin >> k;
  int cur = 2, up = 3, down = 4, size = 4;
  add(cur, up);
  add(cur, down);
  for (int i = 0; i < 35; i++) {
    p2[i] = up;
    int nup = ++size;
    int ndown = ++size;
    add(up, nup);
    add(up, ndown);
    add(down, nup);
    add(down, ndown);
    up = nup;
    down = ndown;
  }
  for (int i = 0; i < 32; i++)
    if ((1 << i) & k) {
      cur = 1;
      for (int t = 0; t < 33 - i; t++) {
        add(cur, ++size);
        cur = size;
      }
      add(cur, p2[i]);
    }
  cout << size << endl;
  for (int i = 1; i <= size; i++) {
    for (int j = 1; j <= size; j++) cout << (ans[i][j] ? 'Y' : 'N');
    cout << endl;
  }
  return 0;
}
