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 Edge[1010][1010];
int Connect1[5010];
int Connect2[5010];
void add(int u, int v) {
  Edge[u][v] = 1;
  Edge[v][u] = 1;
}
int main() {
  int N;
  cin >> N;
  bitset<31> bits(N);
  int node = 2;
  int parent = 0;
  for (int i = 1; i <= 30; i++) {
    int u = node++, v = node++;
    add(parent, v), add(parent, u);
    int z = node++;
    add(v, z), add(u, z);
    Connect1[i] = z;
    parent = z;
  }
  for (int i = 0; i <= 30; i++) {
    Connect2[i] = node;
    add(node, node + 1), node += 1;
    add(node, node + 1), node += 1;
  }
  add(node++, 1);
  for (int i = 0; i < 31; i++) {
    if (bits[i]) {
      add(Connect2[i], Connect1[i]);
    }
  }
  cout << node << endl;
  for (int i = 0; i < node; i++) {
    for (int j = 0; j < node; j++) {
      if (Edge[i][j])
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
  return 0;
}
