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 Maxn = 1000 + 5;
int n;
string s;
bool mark[Maxn][Maxn];
void ok(int l1, int l2, int n1, int n2) {
  for (int i = l1; i <= l2; i++)
    for (int j = n1; j <= n2; j++) mark[i][j] = mark[j][i] = true;
  return;
}
int main() {
  cin >> s;
  n = (int)s.size();
  int lastv = 2, last = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != '0') {
      for (int j = 0; j < i; j++) {
        mark[lastv][last] = mark[last][lastv] = true;
        last = lastv;
        lastv++;
      }
      ok(last, last, lastv, lastv + (s[i] - '0') - 1);
      int l1 = lastv, l2 = lastv + (s[i] - '0') - 1, n1 = lastv + (s[i] - '0'),
          n2 = lastv + (s[i] - '0') + 9;
      for (int j = i; j < n - 1; j++) {
        ok(l1, l2, n1, n2);
        l1 = n1, l2 = n2;
        n1 = l2 + 1, n2 = l2 + 10;
      }
      ok(l1, l2, 1, 1);
      lastv = l2 + 1;
      last = 0;
    }
  }
  cout << lastv << endl;
  for (int i = 0; i < lastv; i++) {
    for (int j = 0; j < lastv; j++) {
      if (mark[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}