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 MAXN = 1005;
bool g[MAXN][MAXN];
void a(int x, int y) { g[x][y] = g[y][x] = true; }
int n;
void out() {
  vector<int> p(n);
  for (int i = 0; i < (int)(n); i++) p[i] = i;
  p[1] = n - 1;
  p[n - 1] = 1;
  cout << n << endl;
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(n); j++) printf("%c", g[p[i]][p[j]] ? 'Y' : 'N');
    cout << endl;
  }
}
int main() {
  int k;
  cin >> k;
  if (k == 1) {
    a(0, 1);
    a(1, 0);
    n = 2;
    out();
    return 0;
  }
  int p = 0;
  while ((1 << p) <= k) p++;
  p--;
  a(0, 1);
  a(0, 2);
  vector<int> cur;
  cur.push_back(1);
  cur.push_back(2);
  n = 3;
  for (int i = 0; i < p; i++) {
    vector<int> ncur;
    a(cur[0], n);
    a(cur[1], n);
    ncur.push_back(n++);
    a(cur[0], n);
    a(cur[1], n);
    ncur.push_back(n++);
    for (int j = 2; j <= cur.size(); j++) {
      a(cur[j - 1], n);
      ncur.push_back(n++);
    }
    cur = ncur;
  }
  for (int i = 0; i < (int)(p + 1); i++) {
    if (k & (1 << i)) {
      a(cur[cur.size() - i - 1], n);
    }
  }
  n++;
  out();
  return 0;
}
