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 = 2e3 + 100;
const long long inf = 1e18 + 10;
int g[maxn][maxn], num[maxn], cnt = 100;
int main() {
  long long b;
  scanf("%lld", &b);
  while (b) {
    num[--cnt] = b % 2;
    b /= 2;
  }
  cnt++;
  long long sum = 2, pst = 1, ped = 1;
  for (int i = cnt; i <= 100; i++) {
    int sst = sum + 1, eed = sum + 2;
    if (i == 100) {
      sst = eed = 2;
    }
    for (int j = pst; j <= ped; j++)
      for (int k = sst; k <= eed; k++) {
        g[j][k] = g[k][j] = 1;
      }
    if (i != 100) sum += 2;
    pst = sst;
    ped = eed;
    if (num[i]) {
      int dian = i - cnt;
      for (int j = sum + 1; j < sum + dian; j++) {
        g[j][j + 1] = g[j + 1][j] = 1;
      }
      if (dian)
        g[1][sum + 1] = g[sum + 1][1] = g[sum][sum + dian] =
            g[sum + dian][sum] = 1,
                   sum += dian;
      else
        g[1][sum + 1] = g[sum + 1][1] = 1, ped++, sum++;
    }
  }
  printf("%lld\n", sum);
  for (int i = 1; i <= sum; i++) {
    for (int j = 1; j <= sum; j++) {
      if (g[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
}