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 n;
char sol[172][172];
void baga(int i, int j) { sol[i][j] = sol[j][i] = 'Y'; }
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  int cn = n, nr = 0, i, j;
  for (i = 1; i <= 170; i++)
    for (j = 1; j <= 170; j++) sol[i][j] = 'N';
  while (cn != 1) {
    nr++;
    cn /= 2;
  }
  int last = 1, u, cu;
  for (i = 1; i <= nr; i++) {
    baga(last, 3 * i);
    baga(last, 3 * i + 1);
    baga(3 * i, 3 * i + 2);
    baga(3 * i + 1, 3 * i + 2);
    last = 3 * i + 2;
  }
  cu = last;
  if (last == 1) last++;
  baga(last - 1, 2);
  baga(last - 2, 2);
  if (last == 2) last--;
  last = 1;
  for (i = 1; i <= nr; i++) {
    baga(last, 100 + 2 * i);
    last = 100 + 2 * i;
    baga(last, last + 1);
    last++;
  }
  last--;
  u = 2;
  while (n != 1) {
    if (n & 1) {
      baga(last, u);
    }
    cu -= 3;
    u = cu;
    last -= 2;
    n /= 2;
  }
  printf("170\n");
  for (i = 1; i <= 170; i++) {
    for (j = 1; j <= 170; j++) printf("%c", sol[i][j]);
    printf("\n");
  }
  return 0;
}
