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;
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
const int N = 1e3 + 10;
const long long MOD = 1e9 + 7;
const int rx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int ry[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const long double eps = 1e-7;
const double pi = acos(-1.0);
const int INF = 2e9;
vector<int> g[N];
int rz[32];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int p = (31 - __builtin_clz(n));
  int z = p - 1;
  g[1].push_back(3);
  int nom = 3;
  for (int i = 0; i < p; i++) {
    if (nom == 3) {
      g[nom].push_back(nom + 1);
      g[nom].push_back(nom + 2);
      nom = 5;
      rz[z] = 4;
      z--;
    } else {
      g[nom - 1].push_back(nom + 1);
      g[nom - 1].push_back(nom + 2);
      g[nom].push_back(nom + 1);
      g[nom].push_back(nom + 2);
      nom += 2;
      rz[z] = nom - 1;
      z--;
    }
  }
  if (!p)
    g[nom].push_back(2);
  else {
    g[nom - 1].push_back(2);
    g[nom].push_back(2);
  }
  nom++;
  for (int i = p - 1; i >= 0; i--) {
    if ((1 & (n >> i))) {
      g[1].push_back(nom);
      for (int j = 0; j < p - i - 1; j++) {
        g[nom].push_back(nom + 1);
        nom++;
      }
      g[nom].push_back(rz[i]);
      nom++;
    }
  }
  cout << nom - 1 << endl;
  vector<vector<char> > pl(nom, vector<char>(nom, 'N'));
  for (int i = 1; i < nom; i++) {
    for (auto it : g[i]) {
      pl[i][it] = 'Y';
      pl[it][i] = 'Y';
    }
  }
  for (int i = 1; i < nom; i++) {
    for (int j = 1; j < nom; j++) cout << pl[i][j];
    cout << endl;
  }
  return 0;
}