Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool g[1010][1010];
int K, keynode[33], single[66];
int cnt = 2;
void modify(int a, int b) { g[a][b] = g[b][a] = 1; }
void getway(int x) {
  int c = 62 - 2 * x - 1;
  modify(single[c], keynode[x]);
}
void work() {
  keynode[0] = 1;
  for (int i = 1; i < 33; i++) {
    modify(cnt, cnt - 1);
    cnt++;
    modify(cnt, cnt - 2);
    cnt++;
    modify(cnt, cnt - 1);
    modify(cnt, cnt - 2);
    keynode[i] = cnt;
    cnt++;
  }
  int pre = 0;
  for (int i = 1; i < 66; i++) {
    modify(pre, cnt);
    single[i] = cnt;
    pre = cnt++;
  }
  for (int i = 0; i < 31; i++)
    if (K >> i & 1) {
      getway(i);
    }
}
int main() {
  scanf("%d", &K);
  work();
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) printf("%c", "NY"[g[i][j] == 1]);
    puts("");
  }
  return 0;
}