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;
int currentN = 0;
vector<vector<int> > g;
vector<vector<int> > back;
int add_vert() {
  g.push_back(vector<int>());
  back.push_back(vector<int>());
  return currentN++;
}
void add_edge(int from, int to) {
  g[from].push_back(to);
  back[to].push_back(from);
}
void copy_income(int from, int to) {
  for (int i = 0; i < (int)back[from].size(); i++) {
    int f = back[from][i];
    add_edge(f, to);
  }
}
int build(int from, int to, int count) {
  if (count == 1) {
    add_edge(from, to);
    return 1;
  } else {
    if (count % 2 == 1) {
      int len = build(from, to, count - 1);
      int last = from;
      for (int i = 0; i < len - 1; i++) {
        int cur = add_vert();
        add_edge(last, cur);
        last = cur;
      }
      add_edge(last, to);
      return len;
    } else {
      int a = add_vert(), b = add_vert();
      add_edge(a, to);
      add_edge(b, to);
      int len = build(from, a, count / 2);
      copy_income(a, b);
      return len + 1;
    }
  }
}
char answer[1001][1001];
int main() {
  ios_base::sync_with_stdio(false);
  int k;
  cin >> k;
  add_vert();
  add_vert();
  build(0, 1, k);
  cout << currentN << endl;
  for (int i = 0; i < currentN; i++) {
    answer[i][currentN] = '\0';
    for (int j = 0; j < currentN; j++) answer[i][j] = 'N';
  }
  for (int i = 0; i < currentN; i++)
    for (int j = 0; j < (int)g[i].size(); j++)
      answer[i][g[i][j]] = answer[g[i][j]][i] = 'Y';
  for (int i = 0; i < currentN; i++) printf("%s\n", answer[i]);
}
