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;
int n;
char G[MAXN][MAXN];
void addedge(int i, int j) {
  G[i][j] = 'Y';
  G[j][i] = 'Y';
}
void solve() {
  if (n == 1) {
    printf("%d\n", 2);
    printf("NY\n");
    printf("YN\n");
    return;
  }
  memset(G, 'N', sizeof G);
  int p = 2, flag = 0, top;
  for (int i = (30); i >= (0); --i)
    if (n & (1 << i)) {
      if (!flag) {
        top = i;
        int t = 1;
        addedge(1, ++p);
        addedge(1, ++p);
        for (int j = (1); j < (i); ++j) {
          t = ++p;
          addedge(p - 2, t);
          addedge(p - 1, t);
          addedge(t, ++p);
          addedge(t, ++p);
        }
        addedge(p - 1, 2);
        addedge(p, 2);
        flag = 1;
      } else {
        int t = 5;
        addedge(1, ++p);
        for (int j = (1); j < (top - i); ++j) {
          addedge(p, p + 1);
          ++p;
          addedge(p, p + 1);
          ++p;
          t += 3;
        }
        if (i == 0)
          addedge(p, 2);
        else
          addedge(p, t);
      }
    }
  printf("%d\n", p);
  for (int i = (1); i <= (p); ++i) {
    for (int j = (1); j <= (p); ++j) printf("%c", G[i][j]);
    printf("\n");
  }
}
int main() {
  while (~scanf("%d", &n)) solve();
  return 0;
}
