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 k, nm;
int v;
int h[1001], d[1001];
int num[30];
bool a[1001][1001];
void add(int x, int y) {
  a[x][y] = 1;
  a[y][x] = 1;
  d[y] += d[x];
  h[y] = h[x] + 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> k;
  nm = 2;
  num[0] = 1;
  d[0] = 1;
  for (int i = 0; i < 29; ++i) {
    add(num[i], ++nm);
    add(num[i], ++nm);
    ++nm;
    add(nm - 2, nm);
    add(nm - 1, nm);
    num[i + 1] = nm;
  }
  for (int i = 29; i >= 0; i--) {
    if (k & (1 << i)) {
      int v = num[i];
      while (h[v] != 58) {
        add(v, ++nm);
        v = nm;
      }
      add(v, 2);
    }
  }
  cout << nm << '\n';
  for (int i = 1; i <= nm; ++i) {
    for (int j = 1; j <= nm; ++j) {
      if (a[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << '\n';
  }
}
