#include <bits/stdc++.h>
using namespace std;
int edge[1005][1005];
long long tpow[35];
void connect(int a, int b) { edge[a][b] = edge[b][a] = 1; }
void init() {
  memset(edge, 0, sizeof(edge));
  connect(1, 3);
  connect(1, 4);
  connect(2, 100);
  for (int i = 3; i <= 61; i += 2) {
    connect(i, i + 2);
    connect(i, i + 3);
    connect(i + 1, i + 2);
    connect(i + 1, i + 3);
  }
  for (int i = 65; i <= 99; i++) connect(i, i + 1);
  tpow[0] = 1;
  for (int i = 1; i <= 32; i++) tpow[i] = tpow[i - 1] * 2;
}
int main() {
  int k;
  while (~scanf("%d", &k)) {
    init();
    for (int i = 32; i > 0; i--)
      if (k >= tpow[i]) {
        connect(i * 2 + 1, 65 + i);
        connect(i * 2 + 2, 65 + i);
        k -= tpow[i];
      }
    if (k) connect(1, 65);
    puts("100");
    for (int i = 1; i <= 100; i++) {
      for (int j = 1; j <= 100; j++)
        if (edge[i][j])
          printf("Y");
        else
          printf("N");
      puts("");
    }
  }
}

### Prompt

Please create a solution in Cpp 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 edge[1005][1005];
long long tpow[35];
void connect(int a, int b) { edge[a][b] = edge[b][a] = 1; }
void init() {
  memset(edge, 0, sizeof(edge));
  connect(1, 3);
  connect(1, 4);
  connect(2, 100);
  for (int i = 3; i <= 61; i += 2) {
    connect(i, i + 2);
    connect(i, i + 3);
    connect(i + 1, i + 2);
    connect(i + 1, i + 3);
  }
  for (int i = 65; i <= 99; i++) connect(i, i + 1);
  tpow[0] = 1;
  for (int i = 1; i <= 32; i++) tpow[i] = tpow[i - 1] * 2;
}
int main() {
  int k;
  while (~scanf("%d", &k)) {
    init();
    for (int i = 32; i > 0; i--)
      if (k >= tpow[i]) {
        connect(i * 2 + 1, 65 + i);
        connect(i * 2 + 2, 65 + i);
        k -= tpow[i];
      }
    if (k) connect(1, 65);
    puts("100");
    for (int i = 1; i <= 100; i++) {
      for (int j = 1; j <= 100; j++)
        if (edge[i][j])
          printf("Y");
        else
          printf("N");
      puts("");
    }
  }
}

```