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 = 1e3 + 1;
char a[maxn][maxn];
int k;
int tot;
inline int newnode() { return tot++; }
int src = 0, dest = 1;
int maxlen;
inline void add(int i, int j) { a[i][j] = a[j][i] = 'Y'; }
void build(int p) {
  vector<int> last, nxt;
  last.push_back(src);
  for (int i = 0; i < p; ++i) {
    nxt = vector<int>({newnode(), newnode()});
    for (int i = 0; i < last.size(); ++i)
      for (int j = 0; j < nxt.size(); ++j) add(last[i], nxt[j]);
    swap(last, nxt);
  }
  for (int i = 0; i < last.size(); ++i) add(last[i], dest);
}
int main() {
  tot = 2;
  for (int i = 0; i < maxn; ++i) {
    fill(a[i], a[i] + maxn, 'N');
  }
  scanf("%d", &k);
  maxlen = log2(k + 0.5);
  for (int i = maxlen; i >= 0; --i) {
    if ((1 << i) & k) {
      build(i);
      k ^= (1 << i);
    }
    if (k) {
      add(src, newnode());
      src = tot - 1;
    }
  }
  for (int i = 0; i < tot; ++i) a[i][tot] = 0;
  printf("%d\n", tot);
  for (int i = 0; i < tot; ++i) puts(a[i]);
}
