Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void PV(vector<long long> v) {
  for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < (long long)v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 1e3 + 10;
long long a[N][N];
map<long long, long long> m;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long k;
  cin >> k;
  long long x = 1, y = 2, u = 1;
  long long l = ceil(log2(k));
  for (long long i = 1; i <= l + 1; i++) {
    m[u] = x;
    y++;
    long long n = y + 2;
    a[x][y]++;
    a[y][x]++;
    a[y][n]++;
    a[n][y]++;
    y++;
    a[x][y]++;
    a[y][x]++;
    a[y][n]++;
    a[n][y]++;
    y++;
    x = n;
    u = u + u;
  }
  y++;
  long long cn = 0;
  long long mx = 0;
  while (u >= 1) {
    long long p = (u);
    if (((p)&k)) {
      mx = max(mx, u);
      cn = max(cn, m[p]);
      long long id = m[p];
      if (id == cn) {
        a[id][2]++;
        a[2][id]++;
        u /= 2;
        continue;
      }
      a[id][y]++;
      a[y][id]++;
      long long s = m[mx] - m[u] - 2 - (m[mx] - m[u] - 1) / 3;
      while (s--) {
        x = y;
        y++;
        a[x][y]++;
        a[y][x]++;
      }
      a[2][y]++;
      a[y][2]++;
      y++;
    }
    u = u / 2;
  }
  cout << (y) << "\n";
  for (long long i = 1; i <= y; i++) {
    for (long long j = 1; j <= y; j++) {
      char c = (a[i][j]) ? 'Y' : 'N';
      cout << c;
    }
    cout << "\n";
  }
  return 0;
}