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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class _T>
inline string tostr(const _T& a) {
  ostringstream os("");
  os << a;
  return os.str();
}
bool grid[1010][1010];
int main() {
  int n;
  while (cin >> n) {
    memset(grid, 0, sizeof(grid));
    int layers = 0;
    int aux = n;
    while (aux) {
      layers++;
      aux = aux >> 1;
    }
    if (layers == 1) {
      printf("2\nNY\nYN\n");
      continue;
    }
    layers--;
    int node = 3;
    grid[1][node] = true;
    grid[1][node + 1] = true;
    for (int(i) = (1); (i) < (layers); ++(i)) {
      grid[node][node + 2] = true;
      grid[node][node + 3] = true;
      grid[node + 1][node + 2] = true;
      grid[node + 1][node + 3] = true;
      node += 2;
    }
    cerr << "node"
         << " = " << node << endl;
    ;
    grid[node][2] = true;
    grid[node + 1][2] = true;
    node += 2;
    for (int(i) = (0); (i) < (layers); ++(i)) {
      if (n & (1 << i)) {
        if (i == 0) {
          grid[1][node] = true;
        } else {
          grid[i * 2 + 1][node] = true;
          grid[i * 2 + 2][node] = true;
        }
        for (int(j) = (i + 1); (j) < (layers); ++(j)) {
          grid[node][node + 1] = true;
          node++;
        }
        grid[node][2] = true;
        node++;
      }
    }
    for (int(i) = (1); (i) < (node + 1); ++(i)) {
      grid[i][i] = false;
      for (int(j) = (1); (j) < (node + 1); ++(j)) {
        if (grid[i][j]) grid[j][i] = true;
      }
    }
    cout << (node - 1) << endl;
    for (int(i) = (1); (i) < (node); ++(i)) {
      for (int(j) = (1); (j) < (node); ++(j)) {
        if (grid[i][j])
          printf("Y");
        else
          printf("N");
      }
      puts("");
    }
  }
  return 0;
}