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 long long maxm = 1e3 + 5;
const long long mod = 998244353;
long long sum[maxm][maxm];
long long d[maxm][maxm];
long long pos[maxm];
long long a[maxm];
long long n, k;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (long long x = 1; x * (k - 1) <= a[n] - a[1]; x++) {
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= k; j++) {
        d[i][j] = 0;
        sum[i][j] = 0;
      }
    }
    for (long long j = 1; j <= k; j++) {
      pos[j] = 0;
    }
    for (long long i = 1; i <= n; i++) {
      d[i][1] = 1;
      sum[i][1] = sum[i - 1][1] + d[i][1];
      for (long long j = 2; j <= k; j++) {
        while (a[i] - a[pos[j] + 1] >= x) {
          pos[j]++;
        }
        d[i][j] = sum[pos[j]][j - 1];
        sum[i][j] = sum[i - 1][j] + d[i][j];
        sum[i][j] %= mod;
      }
    }
    for (long long i = k; i <= n; i++) {
      ans += d[i][k];
      ans %= mod;
    }
  }
  cout << ans << endl;
  return 0;
}
