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.
#include <bits/stdc++.h>
using namespace std;
int a[100], b[1100][1100];
vector<pair<int, int> > ans;
vector<int> c;
void dog(int x, int *a) {
  for (int i = 0; i <= 40; i++) a[i] = 0;
  while (x) {
    a[++a[0]] = (x & 1);
    x = (x >> 1);
  }
}
int main() {
  int n, tot, index, len;
  while (scanf("%d", &n) != EOF) {
    dog(n, a);
    tot = 2;
    ans.clear();
    c.clear();
    for (int i = a[0]; i >= a[0]; i--) {
      if (a[i] == 0) continue;
      index = 1;
      c.push_back(1);
      for (int j = 1; j <= i - 1; j++) {
        ans.push_back(make_pair(index, ++tot));
        ans.push_back(make_pair(index, ++tot));
        ans.push_back(make_pair(tot - 1, tot + 1));
        ans.push_back(make_pair(tot, tot + 1));
        index = ++tot;
        c.push_back(index);
      }
      ans.push_back(make_pair(index, 2));
    }
    for (int i = a[0] - 1; i >= 1; i--) {
      if (a[i] == 0) continue;
      index = c[i - 1];
      for (int j = 2 * i - 1; j <= 2 * (a[0] - 1); j++) {
        ans.push_back(make_pair(index, ++tot));
        index = tot;
      }
      ans.push_back(make_pair(index, 2));
    }
    for (int i = 1; i <= tot; i++)
      for (int j = 1; j <= tot; j++) b[i][j] = 0;
    len = ans.size();
    for (int i = 0; i < len; i++) {
      b[ans[i].first][ans[i].second] = 1;
      b[ans[i].second][ans[i].first] = 1;
    }
    printf("%d\n", tot);
    for (int i = 1; i <= tot; i++) {
      for (int j = 1; j <= tot; j++) printf("%c", (b[i][j] == 0 ? 'N' : 'Y'));
      printf("\n");
    }
  }
  return 0;
}
