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 ch[205][205];
int main() {
  for (int i = 1; i <= 200; i++) {
    for (int j = 1; j <= 200; j++) {
      ch[i][j] = 'N';
    }
  }
  ch[1][3] = 'Y';
  ch[3][1] = 'Y';
  for (int i = 3; i <= 99; i++) {
    if (i % 3 == 0) {
      ch[i][i + 1] = 'Y';
      ch[i + 1][i] = 'Y';
      ch[i][i + 2] = 'Y';
      ch[i + 2][i] = 'Y';
    } else if (i % 3 == 2) {
      ch[i][i + 1] = 'Y';
      ch[i + 1][i] = 'Y';
    } else {
      ch[i][i + 2] = 'Y';
      ch[i + 2][i] = 'Y';
    }
  }
  for (int i = 110; i < 190; i++) {
    ch[i][i + 1] = 'Y';
    ch[i + 1][i] = 'Y';
  }
  ch[190][2] = 'Y';
  ch[2][190] = 'Y';
  int len = 1;
  int dis[32];
  int last = 190;
  int k;
  cin >> k;
  memset(dis, 0, sizeof(dis));
  while (k) {
    if (k & 1) dis[len] = 1;
    k >>= 1;
    len++;
  }
  len--;
  for (int i = 1; i <= len; i++) {
    if (dis[i]) {
      int xx = 2 * (len - i);
      ch[3 * i][last - xx] = 'Y';
      ch[last - xx][3 * i] = 'Y';
    }
  }
  cout << 200 << endl;
  for (int i = 1; i <= 200; i++) {
    for (int j = 1; j <= 200; j++) {
      printf("%c", ch[i][j]);
    }
    printf("\n");
  }
  return 0;
}