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 long long N = 1005;
bool g[N][N];
void add(long long u, long long v) {
  g[u][v] = 1;
  g[v][u] = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long k, cur = 3;
  cin >> k;
  bool nav = 1;
  long long mx = 2;
  vector<pair<long long, long long> > st;
  for (long long i = 30; i >= 0; i--) {
    if ((1ll << i) & k) {
      if (nav) {
        for (long long t = i - 1, atsk = 2; t >= 0; t--, atsk += 2)
          add(mx, mx + 1), add(mx, mx + 2), add(mx + 1, mx + 3),
              add(mx + 2, mx + 3), mx += 3, st.push_back({mx, atsk});
        add(1, mx);
        cur = mx + 1;
        reverse(st.begin(), st.end());
        nav = 0;
      } else {
        auto t = st[i];
        add(2, cur);
        for (long long j = 2; j < t.second; j++) add(cur, cur + 1), cur++;
        add(t.first, cur++);
      }
    }
  }
  cout << cur - 1 << "\n";
  for (long long i = 1; i < cur; i++) {
    for (long long j = 1; j < cur; j++) cout << ((g[i][j]) ? 'Y' : 'N');
    cout << "\n";
  }
  return 0;
}
