#include <bits/stdc++.h>
using namespace std;
const int ms = 180;
const int sigma = 30;
int n, m, k, t = 1000;
int qt = 0;
string ans[ms];
map<int, int> dig;
map<int, int> id;
void createPath() {
  ans[0][2] = 'Y';
  ans[2][0] = 'Y';
  int pos = 2;
  int i = 0;
  int count = 0;
  while (dig.size() < 32) {
    dig[i] = pos;
    count++;
    ans[pos][pos + 1] = 'Y';
    ans[pos][pos + 2] = 'Y';
    ans[pos + 1][pos] = 'Y';
    ans[pos + 2][pos] = 'Y';
    ans[pos + 2][pos + 3] = 'Y';
    ans[pos + 3][pos + 2] = 'Y';
    ans[pos + 1][pos + 3] = 'Y';
    ans[pos + 3][pos + 1] = 'Y';
    pos += 3;
    i++;
  }
  int path = 0;
  i = 100;
  while (path < 32) {
    id[path] = i;
    ans[i][i + 1] = 'Y';
    ans[i + 1][i] = 'Y';
    ans[i + 1][i + 2] = 'Y';
    ans[i + 2][i + 1] = 'Y';
    i += 2;
    path++;
  }
  ans[i][1] = 'Y';
  ans[1][i] = 'Y';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < ms; i++) {
    ans[i] = string(ms, 'N');
  }
  createPath();
  for (int i = 0; i < 32; i++) {
    if (n & (1 << i)) {
      ans[dig[i]][id[i]] = 'Y';
      ans[id[i]][dig[i]] = 'Y';
    }
  }
  cout << ms << endl;
  for (int i = 0; i < ms; i++) {
    cout << ans[i] << endl;
  }
}

### Prompt

Your task is to create a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int ms = 180;
const int sigma = 30;
int n, m, k, t = 1000;
int qt = 0;
string ans[ms];
map<int, int> dig;
map<int, int> id;
void createPath() {
  ans[0][2] = 'Y';
  ans[2][0] = 'Y';
  int pos = 2;
  int i = 0;
  int count = 0;
  while (dig.size() < 32) {
    dig[i] = pos;
    count++;
    ans[pos][pos + 1] = 'Y';
    ans[pos][pos + 2] = 'Y';
    ans[pos + 1][pos] = 'Y';
    ans[pos + 2][pos] = 'Y';
    ans[pos + 2][pos + 3] = 'Y';
    ans[pos + 3][pos + 2] = 'Y';
    ans[pos + 1][pos + 3] = 'Y';
    ans[pos + 3][pos + 1] = 'Y';
    pos += 3;
    i++;
  }
  int path = 0;
  i = 100;
  while (path < 32) {
    id[path] = i;
    ans[i][i + 1] = 'Y';
    ans[i + 1][i] = 'Y';
    ans[i + 1][i + 2] = 'Y';
    ans[i + 2][i + 1] = 'Y';
    i += 2;
    path++;
  }
  ans[i][1] = 'Y';
  ans[1][i] = 'Y';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < ms; i++) {
    ans[i] = string(ms, 'N');
  }
  createPath();
  for (int i = 0; i < 32; i++) {
    if (n & (1 << i)) {
      ans[dig[i]][id[i]] = 'Y';
      ans[id[i]][dig[i]] = 'Y';
    }
  }
  cout << ms << endl;
  for (int i = 0; i < ms; i++) {
    cout << ans[i] << endl;
  }
}

```