#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const int maxn = 2000 + 10;
const int mod = 1000000007;
int k, n;
bool ans[maxn][maxn];
vector<int> bi;
int main() {
  scanf("%d", &k);
  while (k) {
    bi.push_back(k % 2);
    k >>= 1;
  }
  int deep = bi.size() - 1;
  int cnt = 2;
  bool flag = false;
  for (int i = 0; i < bi.size(); i++)
    if (bi[i] == 1) {
      int s = 0, j = 0;
      if (flag) {
        if (i != deep) s = deep - i + 1;
        j = deep - i;
      } else
        for (; j < deep - i; j++) {
          ans[s][cnt] = ans[cnt][s] = true;
          s = cnt;
          cnt++;
          flag = true;
        }
      int s1 = s, s2 = s;
      for (; j < deep; j++) {
        ans[s1][cnt] = ans[cnt][s1] = true;
        ans[s1][cnt + 1] = ans[cnt + 1][s1] = true;
        ans[s2][cnt] = ans[cnt][s2] = true;
        ans[s2][cnt + 1] = ans[cnt + 1][s2] = true;
        s1 = cnt;
        s2 = cnt + 1;
        cnt += 2;
      }
      ans[1][s1] = ans[s1][1] = ans[1][s2] = ans[s2][1] = true;
    }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) {
      if (ans[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    putchar('\n');
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 INF = 1 << 30;
const int maxn = 2000 + 10;
const int mod = 1000000007;
int k, n;
bool ans[maxn][maxn];
vector<int> bi;
int main() {
  scanf("%d", &k);
  while (k) {
    bi.push_back(k % 2);
    k >>= 1;
  }
  int deep = bi.size() - 1;
  int cnt = 2;
  bool flag = false;
  for (int i = 0; i < bi.size(); i++)
    if (bi[i] == 1) {
      int s = 0, j = 0;
      if (flag) {
        if (i != deep) s = deep - i + 1;
        j = deep - i;
      } else
        for (; j < deep - i; j++) {
          ans[s][cnt] = ans[cnt][s] = true;
          s = cnt;
          cnt++;
          flag = true;
        }
      int s1 = s, s2 = s;
      for (; j < deep; j++) {
        ans[s1][cnt] = ans[cnt][s1] = true;
        ans[s1][cnt + 1] = ans[cnt + 1][s1] = true;
        ans[s2][cnt] = ans[cnt][s2] = true;
        ans[s2][cnt + 1] = ans[cnt + 1][s2] = true;
        s1 = cnt;
        s2 = cnt + 1;
        cnt += 2;
      }
      ans[1][s1] = ans[s1][1] = ans[1][s2] = ans[s2][1] = true;
    }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) {
      if (ans[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    putchar('\n');
  }
  return 0;
}

```