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 k, adj[1010][1010];
int main() {
  scanf("%d", &k);
  int node = 2, mx = 0;
  vector<pair<int, int> > res;
  for (int i = 0; k >> i; i++)
    if (k >> i & 1) mx = i;
  if (mx == 0) {
    puts("2\nNY\nYN");
    return 0;
  }
  vector<int> lst;
  for (int i = 0; k >> i; i++) {
    if (k >> i & 1) {
      if (i > 0) {
        res.push_back({0, ++node});
        res.push_back({0, ++node});
        int bef = node - 1;
        for (int j = 1; j < i; j++) {
          res.push_back({bef, ++node});
          res.push_back({bef, ++node});
          res.push_back({bef + 1, node - 1});
          res.push_back({bef + 1, node});
          bef = node - 1;
        }
        if (i < mx) {
          if (lst.empty()) {
            res.push_back({bef, ++node});
            res.push_back({bef + 1, node});
            lst.push_back(node);
            for (int j = i + 1; j < mx; j++) {
              res.push_back({node, node + 1});
              node++;
              lst.push_back(node);
            }
            res.push_back({node, 1});
          } else {
            res.push_back({bef, lst[lst.size() - (mx - i)]});
            res.push_back({bef + 1, lst[lst.size() - (mx - i)]});
          }
        } else {
          res.push_back({bef, 1});
          res.push_back({bef + 1, 1});
        }
      } else {
        int bef = 0;
        for (int j = 0; j < mx; j++) {
          res.push_back({bef, ++node});
          bef = node;
        }
        res.push_back({node, 1});
      }
    }
  }
  printf("%d\n", node + 1);
  for (auto& it : res) adj[it.first][it.second] = adj[it.second][it.first] = 1;
  for (int i = 0; i <= node; i++) {
    for (int j = 0; j <= node; j++)
      if (adj[i][j])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
  return 0;
}