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;
bool edge[1010][1010];
int k;
inline void addEdge(int x, int y) { edge[x][y] = edge[y][x] = 1; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> k;
  int tmp = 1;
  int v = 3;
  int u = v + 2 * 100;
  while (tmp * 2 <= k) {
    addEdge(v, 100 + 2 * v + 0);
    addEdge(v, 100 + 2 * v + 1);
    addEdge(v + 1, 100 + 2 * v + 0);
    addEdge(v + 1, 100 + 2 * v + 1);
    addEdge(u + 0, u + 1);
    addEdge(u + 1, u + 2);
    v++;
    u += 2;
    tmp *= 2;
  }
  addEdge(1, 3 + 2 * 100);
  addEdge(v, 2);
  for (__typeof(30) i = (0); i <= (30); i++)
    if (k & (1 << i)) addEdge(u - (2 * i), v - i);
  cout << 300 << "\n";
  for (__typeof(300) i = (1); i <= (300); i++) {
    for (__typeof(300) j = (1); j <= (300); j++)
      cout << (edge[i][j] ? "Y" : "N");
    cout << "\n";
  }
  return 0;
}