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;
int main() {
  int N;
  cin >> N;
  cout << 120 << endl;
  vector<string> adj(120, string(120, 'N'));
  for (int i = 0; i < 30; ++i) {
    int v = i ? i + 1 : 0;
    int u = i ? i + 2 : 2;
    adj[v][u] = adj[u][v] = 'Y';
    adj[50 + 2 * i][52 + 2 * i] = adj[52 + 2 * i][50 + 2 * i] = 'Y';
    adj[51 + 2 * i][53 + 2 * i] = adj[53 + 2 * i][51 + 2 * i] = 'Y';
    adj[51 + 2 * i][52 + 2 * i] = adj[52 + 2 * i][51 + 2 * i] = 'Y';
    adj[50 + 2 * i][53 + 2 * i] = adj[53 + 2 * i][50 + 2 * i] = 'Y';
  }
  adj[50 + 2 * 29][1] = adj[1][50 + 2 * 29] = 'Y';
  adj[51 + 2 * 29][1] = adj[1][51 + 2 * 29] = 'Y';
  for (int i = 29; i >= 0; i--) {
    int u = i ? i + 2 : 2;
    if (N % 2) {
      adj[50 + 2 * i][u] = adj[u][50 + 2 * i] = 'Y';
    }
    N /= 2;
  }
  for (int i = 0; i < 120; ++i) {
    cout << adj[i] << endl;
  }
}
