#include <bits/stdc++.h>
using namespace std;
int K, N = 1000;
int cnt = 2;
char graph[1005][1005];
int bits[1005];
int main() {
  cin >> K;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      graph[i][j] = 'N';
    }
  }
  if (K == 1) {
    cout << 2 << endl;
    graph[1][2] = graph[2][1] = 'Y';
    cout << graph[1][1] << graph[1][2] << endl;
    cout << graph[1][2] << graph[2][2] << endl;
    return 0;
  }
  int h = 31 - __builtin_clz(K);
  bits[h] = 1;
  for (int i = h - 1; i >= 0; i--) {
    cnt++;
    graph[bits[i + 1]][cnt] = graph[cnt][bits[i + 1]] = 'Y';
    cnt++;
    graph[bits[i + 1]][cnt] = graph[cnt][bits[i + 1]] = 'Y';
    cnt++;
    graph[cnt - 2][cnt] = graph[cnt][cnt - 2] = 'Y';
    graph[cnt - 1][cnt] = graph[cnt][cnt - 1] = 'Y';
    bits[i] = cnt;
    if ((1 << i) & K) {
      int lst = 1;
      for (int k = 0; k < 2 * (h - i) - 1; k++) {
        cnt++;
        graph[lst][cnt] = graph[cnt][lst] = 'Y';
        lst = cnt;
      }
      graph[lst][bits[i]] = graph[bits[i]][lst] = 'Y';
    }
  }
  graph[bits[0]][2] = graph[2][bits[0]] = 'Y';
  cout << cnt << "\n";
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      cout << graph[i][j];
    }
    cout << "\n";
  }
}

### Prompt

Please create a solution in Cpp 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;
int K, N = 1000;
int cnt = 2;
char graph[1005][1005];
int bits[1005];
int main() {
  cin >> K;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      graph[i][j] = 'N';
    }
  }
  if (K == 1) {
    cout << 2 << endl;
    graph[1][2] = graph[2][1] = 'Y';
    cout << graph[1][1] << graph[1][2] << endl;
    cout << graph[1][2] << graph[2][2] << endl;
    return 0;
  }
  int h = 31 - __builtin_clz(K);
  bits[h] = 1;
  for (int i = h - 1; i >= 0; i--) {
    cnt++;
    graph[bits[i + 1]][cnt] = graph[cnt][bits[i + 1]] = 'Y';
    cnt++;
    graph[bits[i + 1]][cnt] = graph[cnt][bits[i + 1]] = 'Y';
    cnt++;
    graph[cnt - 2][cnt] = graph[cnt][cnt - 2] = 'Y';
    graph[cnt - 1][cnt] = graph[cnt][cnt - 1] = 'Y';
    bits[i] = cnt;
    if ((1 << i) & K) {
      int lst = 1;
      for (int k = 0; k < 2 * (h - i) - 1; k++) {
        cnt++;
        graph[lst][cnt] = graph[cnt][lst] = 'Y';
        lst = cnt;
      }
      graph[lst][bits[i]] = graph[bits[i]][lst] = 'Y';
    }
  }
  graph[bits[0]][2] = graph[2][bits[0]] = 'Y';
  cout << cnt << "\n";
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      cout << graph[i][j];
    }
    cout << "\n";
  }
}

```