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.
#include <bits/stdc++.h>
using namespace std;
bool ad[1001][1001];
int c = 2;
void add(int i, int j) { ad[i][j] = ad[j][i] = true; }
int solve(vector<int> tie, int k) {
  pair<int, int> ret;
  int next;
  vector<int> nt;
  if (k == 1) {
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], 1);
    return 0;
  } else if (k % 2 == 0) {
    next = c++;
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], next);
    nt.push_back(next);
    next = c++;
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], next);
    nt.push_back(next);
    return solve(nt, k / 2) + 1;
  } else {
    next = c++;
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], next);
    nt.push_back(next);
    next = c++;
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], next);
    nt.push_back(next);
    int ret = solve(nt, k / 2);
    next = c++;
    for (int i = 0; i < (int)(tie.size()); i++) add(tie[i], next);
    for (int j = 0; j < ret; j++) {
      add(next, c);
      next = c++;
    }
    add(next, 1);
    return ret + 1;
  }
}
int main() {
  int k;
  cin >> k;
  vector<int> in;
  in.push_back(0);
  solve(in, k);
  cout << c << endl;
  for (int i = 0; i < c; i++) {
    for (int j = 0; j < c; j++) {
      if (ad[i][j])
        cout << 'Y';
      else
        cout << 'N';
    }
    cout << endl;
  }
  return 0;
}
