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;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
char grid[1003][1003];
void conn(long long i, long long j) {
  grid[i][j] = 'Y';
  grid[j][i] = 'Y';
}
long long point[100];
int main() {
  long long i, j, k, l, m, n, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  while (cin >> k) {
    for (i = 1; i <= 1000; i++)
      for (j = 1; j <= 1000; j++) grid[i][j] = 'N';
    long long mxbit = 0;
    for (i = 0; i <= 30; i++) point[i] = 0;
    conn(1, 3);
    long long node = 2;
    for (i = 30; i >= 0; i--) {
      if (check(k, i) == 0) continue;
      mxbit = max(mxbit, i);
      node++;
      if (i == mxbit) {
        for (j = 0; j < i; j++) {
          conn(node, node + 1);
          conn(node, node + 2);
          conn(node + 2, node + 3);
          conn(node + 1, node + 3);
          point[j] = node + 1;
          node += 3;
        }
        conn(node, 2);
      } else {
        conn(node, point[i]);
        for (j = i + 1; j < mxbit; j++) {
          conn(node, node + 1);
          conn(node + 1, node + 2);
          node += 2;
        }
        conn(node, 2);
      }
    }
    cout << node << "\n";
    for (i = 1; i <= node; i++) {
      for (j = 1; j <= node; j++) {
        printf("%c", grid[i][j]);
      }
      printf("\n");
    }
  }
}