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;
int g[1000][1000], A[20];
int cnt;
int main() {
  long long int k;
  cin >> k;
  memset(g, 0, sizeof g);
  int sz = 0;
  long long int num = k;
  while (num) {
    sz++;
    num /= 10;
  }
  vector<int> v(sz);
  cnt = 2;
  while (k) {
    long long int val = 1;
    while (val < k) {
      if (val == k) break;
      val *= 10;
    }
    if (val > k) val /= 10;
    if (val == 1) {
      v[0] = cnt;
      A[0] = k;
      cnt += k;
      for (int i = 1; i < sz; i++) {
        v[i] = cnt;
        cnt++;
        A[i] = 1;
      }
      k = 0;
    } else {
      long long int num = 1;
      int idx = 0;
      while (num < val) {
        v[idx] = cnt;
        cnt += 10;
        A[idx] = 10;
        num *= 10;
        idx++;
      }
      v[idx] = cnt;
      A[idx] = k / val;
      idx++;
      cnt += k / val;
      for (; idx < sz; idx++) {
        v[idx] = cnt;
        A[idx] = 1;
        cnt++;
      }
      k = k % val;
    }
    for (int i = v[0]; (i - v[0]) < A[0]; i++) {
      g[i][0] = 1;
      g[0][i] = 1;
    }
    for (int i = v[sz - 1]; (i - v[sz - 1]) < A[sz - 1]; i++) {
      g[i][1] = 1;
      g[1][i] = 1;
    }
    for (int i = 0; i < (sz - 1); i++) {
      for (int a = v[i]; (a - v[i]) < A[i]; a++) {
        for (int b = v[i + 1]; (b - v[i + 1]) < A[i + 1]; b++) {
          g[a][b] = 1;
          g[b][a] = 1;
        }
      }
    }
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) {
      if (g[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}