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 debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const int N = 1505;
const int GG = 31;
int n, w[N][N], k, d[N];
void add(int a, int b) { w[a][b] = w[b][a] = 1; }
void put_edges(int& last, int nr) {
  add(1, last);
  add(1, last + 1);
  last += 2;
  for (int i = 2; i <= nr; i++) {
    add(last, last - 1);
    add(last, last - 2);
    add(last + 1, last - 1);
    add(last + 1, last - 2);
    last += 2;
  }
  add(last - 1, d[GG - nr]);
  add(last - 2, d[GG - nr]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> k;
  int last = 3;
  int p = 2;
  for (int i = 1; i <= GG; i++) {
    add(last, p);
    d[i] = last;
    p = last;
    last++;
  }
  if (k % 2 == 1) add(1, d[GG]);
  for (int i = 1; (1 << i) <= k; i++) {
    if (k & (1 << i)) put_edges(last, i);
  }
  cout << last << '\n';
  for (int i = 1; i <= last; i++) {
    for (int j = 1; j <= last; j++) cout << (w[i][j] ? 'Y' : 'N');
    cout << "\n";
  }
}