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;
const int MAX_N = 1510;
int num;
bool ans[MAX_N][MAX_N];
void connect(int a, int b) { ans[a][b] = ans[b][a] = true; }
int main() {
  connect(1, 3);
  connect(1, 4);
  for (int i = 1; i < 30; i++) {
    connect(i * 2 + 1, i * 2 + 3);
    connect(i * 2 + 1, i * 2 + 4);
    connect(i * 2 + 2, i * 2 + 3);
    connect(i * 2 + 2, i * 2 + 4);
  }
  for (int i = 1; i <= 30; i++) {
    connect(62 + i, 63 + i);
  }
  connect(93, 2);
  scanf("%d", &num);
  int temp = 0;
  while (num > 0) {
    if (num & 1) {
      connect(temp * 2 + 1, 63 + temp);
      if (temp != 0) {
        connect(temp * 2 + 2, 63 + temp);
      }
    }
    temp++;
    num >>= 1;
  }
  printf("%d\n", 93);
  for (int i = 1; i <= 93; i++) {
    for (int j = 1; j <= 93; j++) {
      if (ans[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    putchar('\n');
  }
  return 0;
}
