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;
char g[1000 + 1][1000 + 1];
int INF = INT_MAX / 3;
inline void connect(int v1, int v2) {
  g[v1][v2] = 'Y';
  g[v2][v1] = 'Y';
}
int main() {
  ios_base::sync_with_stdio(false);
  unsigned int k;
  cin >> k;
  int leng = 0;
  for (int i = 0; i < 1000; i++) {
    for (int j = 0; j < 1000; j++) {
      g[i][j] = 'N';
    }
  }
  for (int i = 1; i < 40; i++) {
    if ((k >> i) == 0) {
      leng = i - 1;
      break;
    }
  }
  int n3 = 4;
  while (k != 0) {
    int n1 = 2;
    int n2 = 3;
    n3 = 4;
    int cgoal = 0;
    int kt = k;
    for (int i = 1; i < 32; i++) {
      if ((k >> i) == 0) {
        cgoal = i - 1;
        break;
      }
    }
    k = (k ^ (1 << cgoal));
    if (cgoal != 0) {
      connect(0, n1);
      connect(0, n2);
      for (int i = 1; i < cgoal; i++) {
        connect(n1, n1 + 3);
        connect(n1, n2 + 3);
        connect(n2, n2 + 3);
        connect(n2, n1 + 3);
        n2 += 3;
        n1 += 3;
        n3 += 3;
      }
      connect(n1, n3 + 3);
      connect(n2, n3 + 3);
      n3 += 3;
      for (int i = cgoal; i < leng; i++) {
        connect(n3, n3 + 3);
        n3 += 3;
      }
      connect(n3, 1);
    } else {
      connect(0, n3);
      for (int i = 0; i < leng; i++) {
        connect(n3, n3 + 3);
        n3 += 3;
      }
      connect(n3, 1);
    }
  }
  n3 += 2;
  cout << n3 << endl;
  for (int i = 0; i < n3; i++) {
    g[i][n3] = '\0';
    printf("%s", g[i]);
    putchar('\n');
  }
}