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;
inline string tos(int a) {
  ostringstream second;
  second << a;
  string s = second.str();
  return s;
}
const int maxn = 1000 + 10;
bool o[maxn][maxn];
int n = 2, c = 3, l;
int p = 1, cur, t;
inline void add(int e) {
  ++t;
  if (t == 9)
    cur = 2;
  else {
    cur = c++;
    ++n;
  }
  for (int j = 0; j < e; j++) {
    o[c][p] = o[p][c] = 1;
    o[cur][c] = o[c][cur] = 1;
    ++n;
    ++c;
  }
  p = cur;
}
int main() {
  ios_base::sync_with_stdio(false);
  int k;
  cin >> k;
  string s = tos(k);
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0') continue;
    l = s.size() - 1 - i;
    p = 1;
    t = 0;
    for (int j = 0; j < l; j++) add(10);
    add(s[i] - '0');
    for (int j = 0; j < 8 - l; j++) add(1);
  }
  cout << n << "\n";
  for (int i = 1; i < n + 1; i++) {
    for (int j = 1; j < n + 1; j++) {
      if (o[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
}
