#include <bits/stdc++.h>
using namespace std;
int main() {
  int k;
  cin >> k;
  bool zzzz = (k <= 536870911);
  int cnt = 2, p = 0;
  vector<pair<int, int> > ed;
  vector<int> ind;
  ed.push_back(make_pair(cnt, 1));
  for (int i = 0, pre = cnt++; i < 30 - zzzz; ++i) {
    ind.push_back(pre);
    ed.push_back(make_pair(pre, cnt++));
    ed.push_back(make_pair(pre, cnt++));
    pre = cnt;
    ed.push_back(make_pair(pre, cnt - 1));
    ed.push_back(make_pair(pre, cnt - 2));
    ++cnt;
  }
  for (int i = 0; k; k >>= 1, ++i) {
    if (k & 1) {
      ed.push_back(make_pair(cnt, 0));
      for (int j = 0, pre = cnt++; j < 60 - 2 * (i + zzzz); ++j) {
        ed.push_back(make_pair(cnt, pre));
        pre = cnt++;
      }
      ed.push_back(make_pair(cnt - 1, ind[i]));
    }
  }
  cout << cnt << '\n';
  vector<string> out(cnt, string(cnt, 'N'));
  for (auto e : ed) {
    out[e.first][e.second] = out[e.second][e.first] = 'Y';
  }
  for (int i = 0; i < cnt; ++i) {
    cout << out[i] << '\n';
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int k;
  cin >> k;
  bool zzzz = (k <= 536870911);
  int cnt = 2, p = 0;
  vector<pair<int, int> > ed;
  vector<int> ind;
  ed.push_back(make_pair(cnt, 1));
  for (int i = 0, pre = cnt++; i < 30 - zzzz; ++i) {
    ind.push_back(pre);
    ed.push_back(make_pair(pre, cnt++));
    ed.push_back(make_pair(pre, cnt++));
    pre = cnt;
    ed.push_back(make_pair(pre, cnt - 1));
    ed.push_back(make_pair(pre, cnt - 2));
    ++cnt;
  }
  for (int i = 0; k; k >>= 1, ++i) {
    if (k & 1) {
      ed.push_back(make_pair(cnt, 0));
      for (int j = 0, pre = cnt++; j < 60 - 2 * (i + zzzz); ++j) {
        ed.push_back(make_pair(cnt, pre));
        pre = cnt++;
      }
      ed.push_back(make_pair(cnt - 1, ind[i]));
    }
  }
  cout << cnt << '\n';
  vector<string> out(cnt, string(cnt, 'N'));
  for (auto e : ed) {
    out[e.first][e.second] = out[e.second][e.first] = 'Y';
  }
  for (int i = 0; i < cnt; ++i) {
    cout << out[i] << '\n';
  }
}

```