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;
const int GRAPH_BASIS = -2;
const int GRAPH_LINE = -1;
int x;
int i, j, k, m, n, p, q;
char g[1000][1001];
void add(int x, int y) {
  g[x - 1][y - 1] = 'Y';
  g[y - 1][x - 1] = 'Y';
}
int main() {
  for (i = 0; i < 1000; ++i) {
    for (j = 0; j < 1000; ++j) g[i][j] = 'N';
    g[i][1000] = '\0';
  }
  cin >> k;
  n = 2;
  for (i = 0; i <= 5; ++i) {
    x = k % 64;
    k /= 64;
    if (x > 0) {
      add(1, ++n);
      p = n;
      q = n + x + 1;
      for (j = n + 1; j < q; ++j) {
        add(p, j);
        add(j, q);
      }
      n = q;
      for (j = 0; j < i; ++j) {
        for (p = 1; p <= 4; ++p) add(n, n + p);
        for (p = 1; p <= 4; ++p)
          for (q = 5; q <= 8; ++q) add(n + p, n + q);
        for (p = 5; p <= 8; ++p)
          for (q = 9; q <= 12; ++q) add(n + p, n + q);
        for (q = 9; q <= 12; ++q) add(n + q, n + 13);
        n += 13;
      }
      for (j = i; j < 5; ++j) {
        add(n, n + 1);
        ++n;
        add(n, n + 1);
        ++n;
        add(n, n + 1);
        ++n;
        add(n, n + 1);
        ++n;
      }
      add(n, 2);
    }
  }
  cout << n << endl;
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) cout << g[i][j];
    cout << endl;
  }
}