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;
int dp[1002][1002], sum[1002], ara[1002], po[1002];
int dp_func(int val, int n, int k) {
  po[1] = 1;
  for (int i = 1; i <= n; i++) {
    if (i > 1) po[i] = po[i - 1];
    while (ara[i] - ara[po[i]] >= val && po[i] < i) po[i]++;
  }
  for (int i = 1; i <= n; i++) {
    dp[1][i] = 1;
  }
  for (int j = 2; j <= k; j++) {
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1] + dp[j - 1][i];
      if (sum[i] >= 998244353) sum[i] -= 998244353;
    }
    for (int i = 1; i <= n; i++) {
      dp[j][i] = sum[po[i] - 1];
    }
  }
  long long rem = 0;
  for (int i = 1; i <= n; i++) {
    rem = rem + dp[k][i];
  }
  return rem %= 998244353;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ara[i]);
  }
  sort(ara + 1, ara + 1 + n);
  int sum = 0, ans = 0, g;
  for (g = 0; g * (k - 1) <= ara[n]; g++) {
  }
  for (int i = g - 1; i >= 0; i--) {
    long long d = dp_func(i, n, k);
    d -= sum;
    if (d < 0) d = (d + 998244353) % 998244353;
    ans = (ans + (d * i) % 998244353) % 998244353;
    sum += d;
    sum %= 998244353;
  }
  cout << ans << endl;
}
