#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 7;
int a[maxn][maxn];
void Solve(int p, int &num, int len) {
  a[1][num + 1] = a[1][num + 2] = 1, p >>= 1, len--;
  num += 2;
  while (1) {
    if (p == 1) {
      a[2 + len - 1][num - 1] = a[2 + len - 1][num] = 1;
      break;
    } else {
      a[num - 1][num + 1] = a[num - 1][num + 2] = a[num][num + 1] =
          a[num][num + 2] = 1;
      num += 2;
      p >>= 1;
      len--;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int val = 1, len = 0, num = 2, p = 1;
  while (val <= n) val <<= 1, len++;
  for (int i = 1; i <= len; i++) a[2 + i - 1][2 + i] = 1;
  num += len;
  while (p <= n) {
    if (n & p) {
      if (p == 1)
        a[1][2 + len - 1] = 1;
      else
        Solve(p, num, len);
    }
    p <<= 1;
  }
  for (int i = 1; i < num; i++) {
    for (int j = i + 1; j <= num; j++) {
      if (a[i][j]) a[j][i] = 1;
    }
  }
  printf("%d\n", num);
  for (int i = 1; i <= num; i++) {
    for (int j = 1; j <= num; j++) {
      if (a[i][j])
        printf("Y");
      else
        printf("N");
      if (j == num) printf("\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 maxn = 1e3 + 7;
int a[maxn][maxn];
void Solve(int p, int &num, int len) {
  a[1][num + 1] = a[1][num + 2] = 1, p >>= 1, len--;
  num += 2;
  while (1) {
    if (p == 1) {
      a[2 + len - 1][num - 1] = a[2 + len - 1][num] = 1;
      break;
    } else {
      a[num - 1][num + 1] = a[num - 1][num + 2] = a[num][num + 1] =
          a[num][num + 2] = 1;
      num += 2;
      p >>= 1;
      len--;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int val = 1, len = 0, num = 2, p = 1;
  while (val <= n) val <<= 1, len++;
  for (int i = 1; i <= len; i++) a[2 + i - 1][2 + i] = 1;
  num += len;
  while (p <= n) {
    if (n & p) {
      if (p == 1)
        a[1][2 + len - 1] = 1;
      else
        Solve(p, num, len);
    }
    p <<= 1;
  }
  for (int i = 1; i < num; i++) {
    for (int j = i + 1; j <= num; j++) {
      if (a[i][j]) a[j][i] = 1;
    }
  }
  printf("%d\n", num);
  for (int i = 1; i <= num; i++) {
    for (int j = 1; j <= num; j++) {
      if (a[i][j])
        printf("Y");
      else
        printf("N");
      if (j == num) printf("\n");
    }
  }
  return 0;
}

```