#include <bits/stdc++.h>
using namespace std;
int rec[1005][1005];
int main() {
  int x, p, k, i, j;
  cin >> k;
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  p = log2(k - 1) + 1;
  int n = 3 * p + 2;
  for (x = 1, i = 1; i <= p + 1; i++, x += 3) {
    rec[x][x + 2] = rec[x + 2][x] = 1;
  }
  rec[1][4] = rec[4][1] = 1;
  for (i = 1, x = 3; i <= p; i++, x += 3) {
    rec[x][x + 4] = rec[x + 4][x] = 1;
  }
  rec[1][5] = rec[5][1] = 1;
  for (i = 1, x = 5; i <= p - 1; i++, x += 3) {
    rec[x][x + 3] = rec[x + 3][x] = 1;
  }
  rec[x][2] = rec[2][x] = 1;
  for (i = 3; i <= 3 * p; i++) rec[i][i + 3] = rec[i + 3][i] = 1;
  for (i = 0; i < p; i++) {
    if ((1LL << i) & (k - 1)) {
      int x = (i + 1) * 3 + 1, y = (i + 2) * 3 + 2;
      if (i == p - 1)
        rec[x][2] = rec[2][x] = 1;
      else
        rec[x][y] = rec[y][x] = 1;
    }
  }
  printf("%d\n", n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      printf(rec[i][j] ? "Y" : "N");
    }
    printf("\n");
  }
}

### Prompt

Your challenge is to write 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;
int rec[1005][1005];
int main() {
  int x, p, k, i, j;
  cin >> k;
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  p = log2(k - 1) + 1;
  int n = 3 * p + 2;
  for (x = 1, i = 1; i <= p + 1; i++, x += 3) {
    rec[x][x + 2] = rec[x + 2][x] = 1;
  }
  rec[1][4] = rec[4][1] = 1;
  for (i = 1, x = 3; i <= p; i++, x += 3) {
    rec[x][x + 4] = rec[x + 4][x] = 1;
  }
  rec[1][5] = rec[5][1] = 1;
  for (i = 1, x = 5; i <= p - 1; i++, x += 3) {
    rec[x][x + 3] = rec[x + 3][x] = 1;
  }
  rec[x][2] = rec[2][x] = 1;
  for (i = 3; i <= 3 * p; i++) rec[i][i + 3] = rec[i + 3][i] = 1;
  for (i = 0; i < p; i++) {
    if ((1LL << i) & (k - 1)) {
      int x = (i + 1) * 3 + 1, y = (i + 2) * 3 + 2;
      if (i == p - 1)
        rec[x][2] = rec[2][x] = 1;
      else
        rec[x][y] = rec[y][x] = 1;
    }
  }
  printf("%d\n", n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      printf(rec[i][j] ? "Y" : "N");
    }
    printf("\n");
  }
}

```