#include <bits/stdc++.h>
using namespace std;
bool g[1010][1010];
int bit[32];
int a[100];
int main() {
  int k;
  bit[0] = 1;
  for (int i = 1; i < 31; i++) bit[i] = 2 * bit[i - 1];
  while (cin >> k) {
    memset(g, false, sizeof(g));
    int cnt = 2;
    int Max = -1;
    for (int i = 30; i >= 0; i--)
      if (k & bit[i]) {
        int now = 1;
        int len = 0;
        if (Max == -1) {
          Max = i;
          for (int j = 0; j < i; j++) {
            a[j] = now;
            g[now][cnt + 1] = true;
            g[cnt + 1][now] = true;
            g[now][cnt + 2] = true;
            g[cnt + 2][now] = true;
            g[cnt + 1][cnt + 3] = true;
            g[cnt + 3][cnt + 1] = true;
            g[cnt + 2][cnt + 3] = true;
            g[cnt + 3][cnt + 2] = true;
            now = cnt + 3;
            cnt += 3;
            len += 2;
          }
        } else {
          now = a[i];
          len = 2 * i;
        }
        while (len < 2 * Max) {
          g[now][cnt + 1] = true;
          g[cnt + 1][now] = true;
          now = cnt + 1;
          cnt++;
          len++;
        }
        g[now][2] = 1;
        g[2][now] = 1;
      }
    cout << cnt << endl;
    for (int i = 1; i <= cnt; i++) {
      for (int j = 1; j <= cnt; j++) {
        if (g[i][j])
          printf("Y");
        else
          printf("N");
      }
      cout << endl;
    }
  }
  return 0;
}

### 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;
bool g[1010][1010];
int bit[32];
int a[100];
int main() {
  int k;
  bit[0] = 1;
  for (int i = 1; i < 31; i++) bit[i] = 2 * bit[i - 1];
  while (cin >> k) {
    memset(g, false, sizeof(g));
    int cnt = 2;
    int Max = -1;
    for (int i = 30; i >= 0; i--)
      if (k & bit[i]) {
        int now = 1;
        int len = 0;
        if (Max == -1) {
          Max = i;
          for (int j = 0; j < i; j++) {
            a[j] = now;
            g[now][cnt + 1] = true;
            g[cnt + 1][now] = true;
            g[now][cnt + 2] = true;
            g[cnt + 2][now] = true;
            g[cnt + 1][cnt + 3] = true;
            g[cnt + 3][cnt + 1] = true;
            g[cnt + 2][cnt + 3] = true;
            g[cnt + 3][cnt + 2] = true;
            now = cnt + 3;
            cnt += 3;
            len += 2;
          }
        } else {
          now = a[i];
          len = 2 * i;
        }
        while (len < 2 * Max) {
          g[now][cnt + 1] = true;
          g[cnt + 1][now] = true;
          now = cnt + 1;
          cnt++;
          len++;
        }
        g[now][2] = 1;
        g[2][now] = 1;
      }
    cout << cnt << endl;
    for (int i = 1; i <= cnt; i++) {
      for (int j = 1; j <= cnt; j++) {
        if (g[i][j])
          printf("Y");
        else
          printf("N");
      }
      cout << endl;
    }
  }
  return 0;
}

```