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 ans[1005][1005];
int a[100];
void link(int u, int v) { ans[u][v] = ans[v][u] = 1; }
int s;
void print() {
  for (int i = 1; i <= s; i++) {
    for (int j = 1; j <= s; j++) {
      printf("%c", ans[i][j] == 1 ? 'Y' : 'N');
    }
    printf("\n");
  }
}
int main() {
  long long k;
  cin >> k;
  long long t = k;
  int cnt = 0;
  while (t) {
    a[cnt++] = t & 1;
    t >>= 1;
  }
  int len = 3;
  if (cnt == 1) {
    link(2, 1);
    printf("2\n");
    s = 2;
    print();
    return 0;
  }
  link(2, 3);
  s = 3;
  for (int i = cnt - 2; i >= 0; i--) {
    link(s, s + 1);
    link(s, s + 2);
    int l = s + 1, r = s + 2;
    if (i) {
      s += 3;
      if (a[i]) {
        int t = 2;
        for (int j = 1; j <= len; j++) {
          link(t, s);
          t = s;
          s++;
        }
        s--;
      }
      link(l, s);
      link(r, s);
    } else {
      s += 2;
      if (a[i]) {
        s += 1;
        int t = 2;
        for (int j = 1; j < len; j++) {
          link(t, s);
          t = s;
          s++;
        }
        link(t, 1);
        s--;
      }
      link(l, 1);
      link(r, 1);
    }
    len += 2;
  }
  printf("%d\n", s);
  print();
  return 0;
}
