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>
int mindigit(int n) {
  int resu = 0;
  while (n) {
    resu++;
    n >>= 1;
  }
  return resu;
}
int map[2010][2010];
int main() {
  int tot;
  scanf("%d", &tot);
  int m = mindigit(tot);
  int n = 2;
  memset(map, 0, sizeof(map));
  if (tot == 1) {
    map[1][2] = map[2][1] = 1;
  } else {
    int u = ++n, v = ++n;
    int starter = u;
    for (int i = 1; i <= m - 1; i++) {
      int nu = ++n, nv = ++n;
      map[u][nu] = map[nu][u] = 1;
      map[u][nv] = map[nv][u] = 1;
      map[v][nu] = map[nu][v] = 1;
      map[v][nv] = map[nv][v] = 1;
      u = nu;
      v = nv;
    }
    map[2][u] = map[u][2] = 1;
    map[2][v] = map[v][2] = 1;
    map[1][starter] = map[starter][1] = 1;
    int prev = 1;
    for (int i = m - 2; i >= 0; i--) {
      starter += 2;
      n++;
      map[prev][n] = map[n][prev] = 1;
      prev = n;
      if (tot & (1 << i)) map[starter][prev] = map[prev][starter] = 1;
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      printf("%c", "NY"[map[i][j]]);
    }
    printf("\n");
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}
