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;
const int N = 1005;
vector<int> a[N];
int ans[N][N];
int n, k;
vector<int> q;
int qd[N];
int d = 0;
int main() {
  cin >> k;
  n = 2;
  while (k > 1) {
    if (k % 2 == 1) {
      q.push_back(n);
      qd[n] = d;
      k--;
    } else {
      a[n].push_back(n + 1);
      a[n].push_back(n + 2);
      a[n + 1].push_back(n + 3);
      a[n + 2].push_back(n + 3);
      a[n + 1].push_back(n);
      a[n + 2].push_back(n);
      a[n + 3].push_back(n + 1);
      a[n + 3].push_back(n + 2);
      qd[n] = d;
      qd[n + 1] = d + 1;
      qd[n + 2] = d + 1;
      qd[n + 3] = d + 2;
      d += 2;
      n += 3;
      k /= 2;
    }
  }
  a[n].push_back(1);
  a[1].push_back(n);
  for (int i = 0; i < q.size(); i++) {
    int t = q[i];
    for (int j = qd[t]; j < d; j++) {
      n++;
      qd[n] = qd[t] + 1;
      a[n].push_back(t);
      a[t].push_back(n);
      t = n;
    }
    a[t].push_back(1);
    a[1].push_back(t);
  }
  n++;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < a[i].size(); j++)
      if (a[i][j] == 1)
        a[i][j] = 2;
      else if (a[i][j] == 2)
        a[i][j] = 1;
  swap(a[1], a[2]);
  for (int i = 1; i < n; i++)
    for (int j = 0; j < a[i].size(); j++) ans[i][a[i][j]] = 1;
  cout << (n - 1) << endl;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < n; j++)
      if (ans[i][j])
        printf("Y");
      else
        printf("N");
    cout << endl;
  }
  return 0;
}
