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 k, n = 2, lst[1020];
int to[50];
bool g[1020][1020];
int q[1020], f[1020], d[1020];
void add(int x, int y) { g[x][y] = g[y][x] = true; }
bool check() {
  int i, h, tail, s;
  q[h = tail = 1] = 1;
  f[1] = 1;
  memset(d, -1, sizeof(d));
  d[1] = 0;
  while (h <= tail) {
    s = q[h++];
    for (i = 1; i <= n; i++)
      if (g[s][i]) {
        if (-1 == d[i]) {
          d[i] = d[s] + 1;
          f[i] += f[s];
          q[++tail] = i;
        } else if (d[i] == d[s] + 1)
          f[i] += f[s];
      }
  }
  return f[2] == k;
}
int main() {
  int i, t1, t2, j;
  scanf("%d", &k);
  for (i = 0; i <= 28; i++) {
    to[i] = n;
    t1 = ++n;
    t2 = ++n;
    add(to[i], t1);
    add(to[i], t2);
    ++n;
    add(t1, n);
    add(t2, n);
  }
  to[29] = n;
  for (i = 0; i <= 29; i++)
    if (k >> i & 1) {
      lst[0] = 1;
      for (j = 1; j <= (29 - i) * 2; j++) lst[j] = ++n;
      for (j = 0; j < (29 - i) * 2; j++) add(lst[j], lst[j + 1]);
      add(lst[(29 - i) * 2], to[i]);
    }
  printf("%d\n", n);
  for (i = 1; i <= n; i++, puts(""))
    for (j = 1; j <= n; j++)
      if (g[i][j])
        printf("Y");
      else
        printf("N");
}
