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 = 1100;
char ans[maxn][maxn];
void conn(int x, int y) {
  ans[x][y] = 'Y';
  ans[y][x] = 'Y';
}
int N;
int now = 2;
void dn(int n) {
  for (int i = 0; i < n; i++) {
    conn(now, now + i + 1);
    conn(now + i + 1, now + n + 1);
  }
  now += n + 1;
}
int d3[50];
int main() {
  scanf("%d", &N);
  memset(ans, 'N', sizeof(ans));
  int pp = 0;
  while (N) {
    d3[pp++] = N % 3;
    N /= 3;
  }
  for (int i = 0; i < pp; i++) {
    if (d3[i]) {
      conn(0, now);
      dn(d3[i]);
      for (int j = 0; j < i; j++) dn(3);
      for (int j = i + 1; j < pp; j++) dn(1);
      conn(now, 1);
      now++;
    }
  }
  printf("%d\n", now);
  for (int i = 0; i < now; i++) {
    for (int j = 0; j < now; j++) putchar(ans[i][j]);
    puts("");
  }
  return 0;
}