#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
inline void get(int &x) {
  int f = 1;
  char ch = getchar();
  x = 0;
  while (ch < '0' && ch != '-') ch = getchar();
  if (ch == '-') ch = getchar(), f = -1;
  while (ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
vector<int> v;
int n, l, st, ans, nums, w[101], f[N][N];
void chu(int x) {
  l = nums = 0;
  while (x) {
    w[l] = x % 3;
    if (x % 3) nums++;
    l++, x /= 3;
  }
}
int main() {
  cin >> n;
  chu(n), st = 3;
  for (int i = 0; i <= l - 1; i++)
    if (w[i]) {
      int tt = 0;
      v.clear();
      if (w[i] == 1)
        f[1][st] = 1, v.push_back(st), st++;
      else {
        f[1][st] = 1;
        v.push_back(st++);
        f[1][st] = 1;
        v.push_back(st++);
      }
      for (int j = 0; j <= i; j++) {
        for (int k = 0; k <= ((int)v.size()) - 1; k++)
          f[v[k]][st] = 1, f[v[k]][st + 1] = 1, f[v[k]][st + 2] = 1;
        v.clear(), v.push_back(st);
        v.push_back(st + 1), v.push_back(st + 2), st += 3;
      }
      st -= 3;
      for (int k = 1; k <= 1000; k++) f[k][st + 1] = f[k][st + 2] = 0;
      for (int j = i + 1; j <= l - 1; j++) f[st][st + 1] = 1, st++;
      f[st++][2] = 1;
    }
  cout << --st << endl;
  for (int i = 1; i <= st; i++)
    for (int j = 1; j <= st; j++)
      if (f[i][j]) f[j][i] = 1;
  for (int i = 1; i <= st; i++) {
    for (int j = 1; j <= st; j++)
      if (f[i][j])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 N = 1010;
inline void get(int &x) {
  int f = 1;
  char ch = getchar();
  x = 0;
  while (ch < '0' && ch != '-') ch = getchar();
  if (ch == '-') ch = getchar(), f = -1;
  while (ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
vector<int> v;
int n, l, st, ans, nums, w[101], f[N][N];
void chu(int x) {
  l = nums = 0;
  while (x) {
    w[l] = x % 3;
    if (x % 3) nums++;
    l++, x /= 3;
  }
}
int main() {
  cin >> n;
  chu(n), st = 3;
  for (int i = 0; i <= l - 1; i++)
    if (w[i]) {
      int tt = 0;
      v.clear();
      if (w[i] == 1)
        f[1][st] = 1, v.push_back(st), st++;
      else {
        f[1][st] = 1;
        v.push_back(st++);
        f[1][st] = 1;
        v.push_back(st++);
      }
      for (int j = 0; j <= i; j++) {
        for (int k = 0; k <= ((int)v.size()) - 1; k++)
          f[v[k]][st] = 1, f[v[k]][st + 1] = 1, f[v[k]][st + 2] = 1;
        v.clear(), v.push_back(st);
        v.push_back(st + 1), v.push_back(st + 2), st += 3;
      }
      st -= 3;
      for (int k = 1; k <= 1000; k++) f[k][st + 1] = f[k][st + 2] = 0;
      for (int j = i + 1; j <= l - 1; j++) f[st][st + 1] = 1, st++;
      f[st++][2] = 1;
    }
  cout << --st << endl;
  for (int i = 1; i <= st; i++)
    for (int j = 1; j <= st; j++)
      if (f[i][j]) f[j][i] = 1;
  for (int i = 1; i <= st; i++) {
    for (int j = 1; j <= st; j++)
      if (f[i][j])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
  return 0;
}

```