#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5;
bool vis[N][N];
int main() {
  int m;
  scanf("%d", &m);
  int x = 1, cnt = 0, sz;
  vis[1][3] = vis[3][1] = vis[1][4] = vis[4][1] = 1;
  for (int i = 3;; i += 2) {
    x *= 2;
    cnt++;
    if (x > m) {
      sz = i + 2;
      break;
    }
    vis[i][i + 2] = vis[i][i + 3] = vis[i + 1][i + 2] = vis[i + 1][i + 3] = 1;
    vis[i + 2][i] = vis[i + 3][i] = vis[i + 2][i + 1] = vis[i + 3][i + 1] = 1;
  }
  for (int i = 0; i <= 31; i++) {
    if ((1 << i) & m) {
      int pre = i * 2 + 3;
      for (int j = i + 1; j < cnt; j++) {
        sz++;
        vis[pre][sz] = vis[sz][pre] = 1;
        pre = sz;
      }
      vis[pre][2] = vis[2][pre] = 1;
    }
  }
  printf("%d\n", sz);
  for (int i = 1; i <= sz; i++) {
    for (int j = 1; j <= sz; j++) printf("%c", vis[i][j] ? 'Y' : 'N');
    puts("");
  }
  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 N = 1e3 + 5;
bool vis[N][N];
int main() {
  int m;
  scanf("%d", &m);
  int x = 1, cnt = 0, sz;
  vis[1][3] = vis[3][1] = vis[1][4] = vis[4][1] = 1;
  for (int i = 3;; i += 2) {
    x *= 2;
    cnt++;
    if (x > m) {
      sz = i + 2;
      break;
    }
    vis[i][i + 2] = vis[i][i + 3] = vis[i + 1][i + 2] = vis[i + 1][i + 3] = 1;
    vis[i + 2][i] = vis[i + 3][i] = vis[i + 2][i + 1] = vis[i + 3][i + 1] = 1;
  }
  for (int i = 0; i <= 31; i++) {
    if ((1 << i) & m) {
      int pre = i * 2 + 3;
      for (int j = i + 1; j < cnt; j++) {
        sz++;
        vis[pre][sz] = vis[sz][pre] = 1;
        pre = sz;
      }
      vis[pre][2] = vis[2][pre] = 1;
    }
  }
  printf("%d\n", sz);
  for (int i = 1; i <= sz; i++) {
    for (int j = 1; j <= sz; j++) printf("%c", vis[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}

```