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 + 10;
bool a[maxn][maxn];
void addEdge(int i, int j) { a[i][j] = a[j][i] = true; }
int main() {
  int n;
  vector<int> v;
  while (~scanf("%d", &n)) {
    memset(a, 0, sizeof(a));
    v.clear();
    while (n) {
      v.push_back(n % 2);
      n /= 2;
    }
    int temp = 0;
    for (int i = v.size() - 1; i >= 0; i--) {
      if (v[i]) {
        temp = i;
        break;
      }
    }
    addEdge(1, 3);
    addEdge(1, 4);
    int num1, num2;
    num1 = num2 = 3;
    for (int i = 2; i < temp + 2; i++) {
      for (int j = 0; j < i; j++) {
        addEdge(num1 + j, num1 + j + i);
        addEdge(num1 + j, num1 + i + i);
      }
      num1 += i;
      num2 = num1;
    }
    for (int i = 0; i <= temp; i++) {
      if (v[i]) {
        addEdge(num2 + i + 1, 2);
      }
    }
    printf("%d\n", num2 + temp + 1);
    for (int i = 1; i <= num2 + temp + 1; i++) {
      for (int j = 1; j <= num2 + temp + 1; j++) {
        if (a[i][j]) {
          printf("Y");
        } else {
          printf("N");
        }
      }
      printf("\n");
    }
  }
  return 0;
}