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;
string s[1000];
int main() {
  int k;
  cin >> k;
  for (int i = 0; i < 1000; i++) {
    s[i] = string(1000, 'N');
    for (int j = 0; j < 1000; j++) {
      s[i][j] = 'N';
    }
  }
  int next = 2;
  while (k > 0) {
    int x = 1;
    bool four = false;
    if (k < 500) four = true;
    if (four) {
      while (x * x * x * x <= k) x++;
    } else {
      while (x * x * x * x * x <= k) x++;
    }
    x--;
    if (four) {
      k -= x * x * x * x;
    } else {
      k -= x * x * x * x * x;
    }
    vector<int> v;
    for (int i = 0; i < x; i++) {
      s[0][next] = 'Y';
      s[next][0] = 'Y';
      v.push_back(next);
      next++;
    }
    vector<int> temp;
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < v.size(); j++) {
        s[v[j]][next] = 'Y';
        s[next][v[j]] = 'Y';
      }
      temp.push_back(next);
      next++;
    }
    for (int i = 0; i < v.size(); i++) v[i] = temp[i];
    temp.clear();
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < v.size(); j++) {
        s[v[j]][next] = 'Y';
        s[next][v[j]] = 'Y';
      }
      temp.push_back(next);
      next++;
    }
    for (int i = 0; i < v.size(); i++) v[i] = temp[i];
    temp.clear();
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < v.size(); j++) {
        s[v[j]][next] = 'Y';
        s[next][v[j]] = 'Y';
      }
      temp.push_back(next);
      next++;
    }
    for (int i = 0; i < v.size(); i++) v[i] = temp[i];
    temp.clear();
    temp.resize(0);
    if (four) {
      for (int i = 0; i < x; i++) {
        s[v[i]][next] = 'Y';
        s[next][v[i]] = 'Y';
        temp.push_back(next);
        next++;
      }
    } else {
      for (int i = 0; i < x; i++) {
        for (int j = 0; j < v.size(); j++) {
          s[v[j]][next] = 'Y';
          s[next][v[j]] = 'Y';
        }
        temp.push_back(next);
        next++;
      }
    }
    if (v.size() != temp.size()) {
      cout << "WTF"
           << " " << v.size() << " " << temp.size() << endl;
    }
    for (int j = 0; j < temp.size(); j++) {
      s[temp[j]][1] = 'Y';
      s[1][temp[j]] = 'Y';
    }
  }
  cout << next << endl;
  for (int i = 0; i < next; i++) {
    for (int j = 0; j < next; j++) cout << s[i][j];
    cout << endl;
  }
  return 0;
}