Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
bool g[maxn][maxn];
inline void addedge(int u, int v) { g[u][v] = g[v][u] = 1; }
int main() {
  int n;
  scanf("%d", &n);
  n--;
  int tmp = n, lay = 0;
  while (tmp > 0) lay++, tmp /= 2;
  if (lay == 0) {
    puts("2\nNY\nYN");
    exit(0);
  }
  addedge(1, 3);
  addedge(1, 4);
  addedge(1, 5);
  addedge(lay * 3, 2);
  for (int i = 1; i < lay; i++) {
    addedge(i * 3, (i + 1) * 3);
    addedge(i * 3 + 1, (i + 1) * 3 + 1);
    addedge(i * 3 + 1, (i + 1) * 3 + 2);
    addedge(i * 3 + 2, (i + 1) * 3 + 2);
    addedge(i * 3 + 2, (i + 1) * 3 + 1);
  }
  int now = 0;
  while (n > 0) {
    now++;
    if (n & 1) {
      if (now == lay)
        addedge(2, 3 * now + 1);
      else
        addedge(3 * (now + 1), 3 * now + 1);
    }
    n >>= 1;
  }
  int ans = lay * 3 + 2;
  printf("%d\n", ans);
  for (int i = 1; i <= ans; i++) {
    for (int j = 1; j <= ans; j++) {
      if (g[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
  return 0;
}