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;
using ll = long long;
using ld = long double;
using ii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vi>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int INF = 2000000000;
const ll LLINF = 9000000000000000000;
int main() {
  ll k;
  cin >> k;
  if (k == 0) {
    cout << 2 << endl << "NN" << endl << "NN" << endl;
    return 0;
  }
  int vibt = 0;
  ll c = 1;
  while (c < k) {
    c *= 2;
    vibt++;
  }
  int V = 2;
  vvi adj(2, vi());
  vi pleft(vibt + 3, -1);
  for (ll m = 1, d = 0; k > 0; ++d, m *= 2) {
    if ((k & m) == 0LL) continue;
    int ll = 0, lr = -1;
    bool joined = false;
    for (int l = 0; l < vibt; ++l) {
      if (l < d) {
        adj.push_back(vi());
        adj.push_back(vi());
        adj[ll].push_back(V);
        adj[ll].push_back(V + 1);
        if (lr != -1) {
          adj[lr].push_back(V);
          adj[lr].push_back(V + 1);
        }
        ll = V;
        lr = V + 1;
        V += 2;
      } else {
        if (pleft[l] != -1) {
          adj[ll].push_back(pleft[l]);
          if (lr != -1) adj[lr].push_back(pleft[l]);
          joined = true;
          break;
        } else {
          adj.push_back(vi());
          adj[ll].push_back(V);
          if (lr != -1) adj[lr].push_back(V);
          ll = V;
          lr = -1;
          pleft[l] = V;
          ++V;
        }
      }
    }
    if (!joined) {
      adj[ll].push_back(1);
      if (lr != -1) adj[lr].push_back(1);
    }
    k -= m;
  }
  vector<vb> mat(V, vb(V, false));
  for (int i = 0; i < V; ++i)
    for (int j : adj[i]) mat[i][j] = mat[j][i] = true;
  cout << V << '\n';
  for (int i = 0; i < V; ++i) {
    for (int j = 0; j < V; ++j) cout << (mat[i][j] ? 'Y' : 'N');
    cout << '\n';
  }
  cout << flush;
  return 0;
}