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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
const int N = 2005;
int k, n = 2, S = 1, T = 2, bh[33][4];
bool e[N][N];
void add(int u, int v) { e[u][v] = e[v][u] = 1; }
int main() {
  read(k);
  add(S, bh[0][1] = ++n);
  add(S, bh[0][2] = ++n);
  add(S, bh[0][3] = ++n);
  for (register int i = 1; i <= 30; i++) {
    bh[i][1] = ++n;
    bh[i][2] = ++n;
    bh[i][3] = ++n;
    for (register int j = 1; j <= 3; j++)
      add(bh[i - 1][1], bh[i][j]), add(bh[i - 1][2], bh[i][j]);
  }
  for (register int bit = 0; bit <= 30; bit++)
    if ((k >> bit) & 1) {
      int lst = bh[bit][3];
      for (register int i = bit + 1; i <= 30; i++) {
        add(lst, ++n);
        lst = n;
      }
      add(lst, T);
    }
  printf("%d\n", n);
  for (register int i = 1; i <= n; i++) {
    for (register int j = 1; j <= n; j++) printf("%c", e[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}