#include <bits/stdc++.h>
using namespace std;
int cur;
bool c[1011][1011];
int x[33], y[33], z[33];
void connect(int target, int between) {
  for (int i = (1), _b = (between); i <= _b; i++) z[i] = ++cur;
  z[between + 1] = target;
  z[0] = 1;
  for (int i = (0), _b = (between); i <= _b; i++) c[z[i]][z[i + 1]] = true;
}
int main() {
  int n;
  while (cin >> n) {
    memset(c, 0, sizeof c);
    cur = 2;
    for (int i = (1), _b = (30); i <= _b; i++) x[i] = ++cur;
    for (int i = (1), _b = (30); i <= _b; i++) y[i] = ++cur;
    c[x[30]][2] = c[y[30]][2] = true;
    for (int i = (1), _b = (29); i <= _b; i++) {
      c[x[i]][x[i + 1]] = true;
      c[y[i]][y[i + 1]] = true;
      c[y[i]][x[i + 1]] = true;
      c[x[i]][y[i + 1]] = true;
    }
    for (int i = (29), _b = (0); i >= _b; i--)
      if ((1 << i) <= n) {
        n -= 1 << i;
        connect(x[30 - i], 30 - i - 1);
      }
    cout << cur << endl;
    for (int i = (1), _b = (cur); i <= _b; i++) {
      for (int j = (1), _b = (cur); j <= _b; j++)
        if (c[i][j] || c[j][i])
          putchar('Y');
        else
          putchar('N');
      puts("");
    }
  }
  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;
int cur;
bool c[1011][1011];
int x[33], y[33], z[33];
void connect(int target, int between) {
  for (int i = (1), _b = (between); i <= _b; i++) z[i] = ++cur;
  z[between + 1] = target;
  z[0] = 1;
  for (int i = (0), _b = (between); i <= _b; i++) c[z[i]][z[i + 1]] = true;
}
int main() {
  int n;
  while (cin >> n) {
    memset(c, 0, sizeof c);
    cur = 2;
    for (int i = (1), _b = (30); i <= _b; i++) x[i] = ++cur;
    for (int i = (1), _b = (30); i <= _b; i++) y[i] = ++cur;
    c[x[30]][2] = c[y[30]][2] = true;
    for (int i = (1), _b = (29); i <= _b; i++) {
      c[x[i]][x[i + 1]] = true;
      c[y[i]][y[i + 1]] = true;
      c[y[i]][x[i + 1]] = true;
      c[x[i]][y[i + 1]] = true;
    }
    for (int i = (29), _b = (0); i >= _b; i--)
      if ((1 << i) <= n) {
        n -= 1 << i;
        connect(x[30 - i], 30 - i - 1);
      }
    cout << cur << endl;
    for (int i = (1), _b = (cur); i <= _b; i++) {
      for (int j = (1), _b = (cur); j <= _b; j++)
        if (c[i][j] || c[j][i])
          putchar('Y');
        else
          putchar('N');
      puts("");
    }
  }
  return 0;
}

```