Let's call beauty of an array b_1, b_2, …, b_n (n > 1) — min_{1 ≤ i < j ≤ n} |b_i - b_j|.

You're given an array a_1, a_2, … a_n and a number k. Calculate the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

Input

The first line contains integers n, k (2 ≤ k ≤ n ≤ 1000).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^5).

Output

Output one integer — the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

Examples

Input


4 3
1 7 3 5


Output


8

Input


5 5
1 10 100 1000 10000


Output


9

Note

In the first example, there are 4 subsequences of length 3 — [1, 7, 3], [1, 3, 5], [7, 3, 5], [1, 7, 5], each of which has beauty 2, so answer is 8.

In the second example, there is only one subsequence of length 5 — the whole array, which has the beauty equal to |10-1| = 9.
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, K;
int a[2020];
long long f[2020][2020], sum[2020][2020];
inline long long solve(int t) {
  for (int i = 1; i <= K; i++)
    for (int j = 1; j <= n; j++) f[i][j] = sum[i][j] = 0;
  for (int i = 1; i <= n; i++) f[1][i] = 1;
  for (int i = 1; i <= K; i++) {
    int k = 1;
    for (int j = 1; j <= n; j++) {
      if (j < i) continue;
      while (a[j] - a[k] >= t) k++;
      (f[i][j] += sum[i - 1][k - 1]) %= 998244353;
      sum[i][j] = (sum[i][j - 1] + f[i][j]) % 998244353;
    }
  }
  long long re = 0;
  for (int i = 1; i <= n; i++) (re += f[K][i]) %= 998244353;
  return re;
}
int main() {
  n = read(), K = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= a[n] / (K - 1); i++) {
    (ans += solve(i) % 998244353) %= 998244353;
  }
  printf("%I64d", ans);
  return 0;
}
