#include <bits/stdc++.h>
using namespace std;
int g[111][111];
int main() {
  memset(g, 0, sizeof g);
  g[1][3] = g[3][1] = 1;
  g[1][4] = g[4][1] = 1;
  g[2][100] = g[100][2] = 1;
  for (int i = 3; i < 58; i += 2) {
    g[i][i + 2] = g[i + 2][i] = 1;
    g[i + 3][i] = g[i][i + 3] = 1;
    g[i + 1][i + 2] = g[i + 2][i + 1] = 1;
    g[i + 1][i + 3] = g[i + 3][i + 1] = 1;
  }
  for (int i = 61; i < 100; i++) {
    g[i][i + 1] = g[i + 1][i] = 1;
  }
  int n;
  scanf("%d", &n);
  int tmp = n;
  for (int i = 30; i > 0; i--) {
    if (tmp - (1 << i) >= 0) {
      tmp -= (1 << i);
      g[2 * i + 1][61 + i] = g[61 + i][2 * i + 1] = 1;
      g[2 * i + 2][61 + i] = g[61 + i][2 * i + 2] = 1;
    }
  }
  if (tmp > 0) {
    g[1][61] = g[61][1] = 1;
  }
  printf("100\n");
  for (int i = 1; i <= 100; i++) {
    for (int j = 1; j <= 100; j++) {
      if (g[i][j]) {
        printf("Y");
      } else {
        printf("N");
      }
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 g[111][111];
int main() {
  memset(g, 0, sizeof g);
  g[1][3] = g[3][1] = 1;
  g[1][4] = g[4][1] = 1;
  g[2][100] = g[100][2] = 1;
  for (int i = 3; i < 58; i += 2) {
    g[i][i + 2] = g[i + 2][i] = 1;
    g[i + 3][i] = g[i][i + 3] = 1;
    g[i + 1][i + 2] = g[i + 2][i + 1] = 1;
    g[i + 1][i + 3] = g[i + 3][i + 1] = 1;
  }
  for (int i = 61; i < 100; i++) {
    g[i][i + 1] = g[i + 1][i] = 1;
  }
  int n;
  scanf("%d", &n);
  int tmp = n;
  for (int i = 30; i > 0; i--) {
    if (tmp - (1 << i) >= 0) {
      tmp -= (1 << i);
      g[2 * i + 1][61 + i] = g[61 + i][2 * i + 1] = 1;
      g[2 * i + 2][61 + i] = g[61 + i][2 * i + 2] = 1;
    }
  }
  if (tmp > 0) {
    g[1][61] = g[61][1] = 1;
  }
  printf("100\n");
  for (int i = 1; i <= 100; i++) {
    for (int j = 1; j <= 100; j++) {
      if (g[i][j]) {
        printf("Y");
      } else {
        printf("N");
      }
    }
    printf("\n");
  }
  return 0;
}

```