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>
inline int rd() {
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  int x = c - '0';
  while (isdigit(c = getchar())) x = x * 10 + c - '0';
  return x;
}
int k, n, g[1001][1001], pos[32];
void input() { k = rd(); }
inline void ins(int u, int v) {
  if (u == v) return;
  g[u][v] = 1, g[v][u] = 1;
}
void solve() {
  n = 2;
  pos[1] = n = 3;
  for (int i = 2, _ = 30; i <= _; i++) pos[i] = ++n, ins(n - 1, n);
  ins(n, 2);
  for (int i = 0, _ = 30; i <= _; i++)
    if (k >> i & 1) {
      int p1 = 1, p2 = 1;
      for (int j = 1, _ = i; j <= _; j++) {
        n += 2;
        ins(p1, n - 1);
        ins(p1, n);
        ins(p2, n - 1);
        ins(p2, n);
        p1 = n - 1;
        p2 = n;
      }
      ins(p1, pos[i + 1]);
      ins(p2, pos[i + 1]);
    }
  printf("%d\n", n);
  for (int i = 1, _ = n; i <= _; i++) {
    for (int j = 1, _ = n; j <= _; j++) putchar(g[i][j] ? 'Y' : 'N');
    puts("");
  }
}
int main() {
  input();
  solve();
  return 0;
}
