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>
char a[1005][1005];
int dist[1005];
inline void setEdge(int x, int y) { a[x][y] = a[y][x] = 'Y'; }
int main() {
  int k, n = 100;
  scanf("%d", &k);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) a[i][j] = 'N';
  for (int i = 1; i <= 30; i++) setEdge(i, i + 1);
  for (int i = 1; i < 30; i++) {
    setEdge(31 + i, 31 + i + 1);
    setEdge(31 + i, 61 + i + 1);
    setEdge(61 + i, 31 + i + 1);
    setEdge(61 + i, 61 + i + 1);
  }
  setEdge(0, 61);
  setEdge(0, 91);
  for (int i = 0; i < 30; i++)
    if (k & (1 << i)) setEdge(61 - i, 30 - i);
  printf("%d\n", n);
  for (int i = 0; i < n; i++) puts(a[i]);
  return 0;
}
