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 z, n, c;
string k;
vector<int> f;
vector<int> l[1007];
int tab[1007][1007];
void solve(int a, int x) {
  for (int i = 0; i <= k.size(); i++) l[i].clear();
  for (int i = 0; i < int(a); i++) l[0].push_back(c++);
  for (int i = 1; i <= k.size() - 1 - x; i++) l[i].push_back(c++);
  for (int i = k.size() - x; i <= k.size() - 1; i++) {
    for (int j = 0; j < int(10); j++) {
      l[i].push_back(c++);
    }
  }
  for (int i = 1; i <= k.size() - 1; i++) {
    for (int j = 0; j < int(l[i - 1].size()); j++) {
      for (int k = 0; k < int(l[i].size()); k++) {
        tab[l[i - 1][j]][l[i][k]] = 1;
        tab[l[i][k]][l[i - 1][j]] = 1;
      }
    }
  }
  for (int i = 0; i < int(l[0].size()); i++)
    tab[0][l[0][i]] = tab[l[0][i]][0] = 1;
  for (int i = 0; i < int(l[k.size() - 1].size()); i++)
    tab[l[k.size() - 1][i]][1] = tab[1][l[k.size() - 1][i]] = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> k;
  c = 2;
  for (int i = 0; i < int(k.size()); i++) {
    solve(k[i] - '0', k.size() - 1 - i);
  }
  cout << c << endl;
  cerr << c << endl;
  for (int i = 0; i < int(c); i++) {
    for (int j = 0; j < int(c); j++) cout << (tab[i][j] ? "Y" : "N");
    cout << endl;
  }
  return 0;
}
