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 inf = (long long)2e9;
const long long md = 1000000007;
long long dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
long long power(long long a, long long p, long long md) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans *= a;
    p /= 2;
    a *= a;
    ans %= md;
    a %= md;
  }
  return ans;
}
long long mdinv(long long a) { return power(a, md - 2, md); }
long long arr[1003];
long long n, k;
long long DP[1003][1003];
long long last[1003];
signed main() {
  const long long mm = 998244353;
  cin >> n >> k;
  for (long long i = 0; i < (long long)n; i++) cin >> arr[i];
  sort(arr, arr + n);
  long long ans = 0;
  for (long long diff = 1; diff <= (arr[n - 1]) / (k - 1) + 2; diff++) {
    long long first = 0, second = 0;
    for (; second < n; second++) {
      while (arr[second] - arr[first] >= diff) first++;
      last[second] = first - 1;
    }
    for (long long i = 0; i < (long long)1003; i++) DP[0][i] = 0;
    DP[0][1] = 1;
    for (long long i = 1; i < n; i++) {
      DP[i][1] = i + 1;
      for (long long len = 2; len <= k; len++) {
        DP[i][len] = DP[i - 1][len];
        if (last[i] >= 0) DP[i][len] += DP[last[i]][len - 1];
        DP[i][len] %= mm;
      }
    }
    ans += DP[n - 1][k];
    ans %= mm;
  }
  cout << ans;
}
