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;
int mdist;
int n;
bool g[1111][1111];
void edge(int a, int b) { g[a][b] = g[b][a] = true; }
void path(int a, int b, int dist) {
  for (int i = dist; i < mdist; i++) {
    edge(a, n);
    a = n;
    n++;
  }
  edge(a, b);
}
void go(int beg, int end, int k, int dist) {
  if (k & 1) {
    path(beg, end, dist);
  }
  if (k > 1) {
    edge(beg, n);
    edge(beg, n + 1);
    edge(n, n + 2);
    edge(n + 1, n + 2);
    n += 3;
    go(n - 1, end, k / 2, dist + 2);
  }
}
int main() {
  int k;
  scanf("%d", &k);
  n = 2;
  int K = k;
  while (K > 1) {
    mdist += 2;
    K /= 2;
  }
  go(0, 1, k, 0);
  printf("%d\n", n);
  for (int i = (0); i <= (int(n) - 1); ++i) {
    for (int j = (0); j <= (int(n) - 1); ++j) printf("%c", g[i][j] ? 'Y' : 'N');
    printf("\n");
  }
  return 0;
}