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;
vector<int> graph[1010];
int mat[1010][1010];
int nxt;
int main(void) {
  int k;
  scanf("%d", &k);
  int src = 1, dest = 2;
  nxt = 3;
  int power = 0;
  while (k > 0) {
    int dig = k % 10;
    k /= 10;
    if (dig == 0) {
      power++;
      continue;
    } else {
      int start = nxt;
      vector<int> prev;
      for (int i = 0; i < dig; i++) {
        graph[src].push_back(nxt++);
        graph[nxt - 1].push_back(src);
        prev.push_back(nxt - 1);
      }
      int cnt = 0;
      for (int i = 0; i < 10; i++, cnt++) {
        vector<int> v;
        int nodes = cnt < power ? 10 : 1;
        for (int j = 0; j < nodes; j++) {
          for (int x : prev) {
            graph[x].push_back(nxt);
            graph[nxt].push_back(x);
          }
          v.push_back(nxt);
          nxt++;
        }
        prev = v;
      }
      assert(prev.size() == 1);
      for (int x : prev) {
        graph[dest].push_back(x);
        graph[x].push_back(dest);
      }
    }
    power++;
  }
  int fin = nxt - 1;
  printf("%d\n", fin);
  for (int i = 1; i <= fin; i++) {
    for (int x : graph[i]) {
      mat[i][x] = 1;
    }
    for (int j = 1; j <= fin; j++) printf("%s", mat[i][j] ? "Y" : "N");
    printf("\n");
  }
  return 0;
}