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 MOD = 998244353;
const long double EPS = 0.0000001;
long long pows(long long a, long long b) {
  if (b == 0) return 1LL;
  long long temp = pows(a, b / 2);
  temp = (temp * temp) % MOD;
  if (b % 2) temp = (temp * a) % MOD;
  return temp;
}
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return (a - b + MOD) % MOD; }
const int N = 1010;
long long dp[N][N];
long long pref[N][N];
long long arr[N];
int n, k;
long long solve(int what) {
  int lst[n + 1];
  int prev = 0;
  for (int i = 1; i <= n; i++) {
    while (arr[i] - arr[prev + 1] >= what) prev++;
    lst[i] = prev;
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[j][i] = 0;
      pref[j][i] = 0;
    }
  }
  pref[0][0] = dp[0][0] = 1;
  for (int i = 0; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      pref[i][j] = pref[i][j - 1];
      if (i != 0) {
        dp[i][j] = pref[i - 1][lst[j]];
      }
      pref[i][j] = add(dp[i][j], pref[i][j]);
    }
  }
  return pref[k][n];
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  arr[0] = 0;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  sort(arr, arr + n + 1);
  int go_till = arr[n] / (k - 1);
  long long ans = 0;
  for (int i = 1; i <= go_till; i++) {
    ans = add(ans, solve(i));
  }
  cout << ans << endl;
  return 0;
}
