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;
const int SIZEN = 1010;
const int mod = 998244353;
int N, K;
long long a[SIZEN] = {0};
long long ans = 0;
long long f[SIZEN][SIZEN] = {0};
long long p[SIZEN] = {0};
int main() {
  long long Max = 0;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%lld", &a[i]), Max = max(Max, a[i]);
  Max = Max / (K - 1);
  sort(a + 1, a + 1 + N);
  a[0] = -0x3f3f3f3f;
  for (int x = 1; x <= Max; x++) {
    f[0][0] = 1;
    for (int j = 1; j <= K; j++) {
      int now = -1;
      long long sum = 0;
      for (int i = j; i <= N; i++) {
        while (now + 1 < i && a[i] - a[now + 1] >= x) {
          now++;
          sum += f[j - 1][now];
          sum %= mod;
        }
        f[j][i] = sum;
        f[j][i] %= mod;
        if (j == K) {
          ans += f[j][i];
          ans %= mod;
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
