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;
long long n, k;
long long a[1200];
int dp[1200][1200];
int s[1200], now[1200];
long long tot[120000];
long long ans;
long long solve(long long x) {
  int i, j, ret;
  for (i = 0; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      dp[i][j] = 0;
    }
    s[i] = now[i] = 0;
  }
  dp[0][0] = 1;
  s[0] = 1;
  now[0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < k && j < i; j++) {
      while (now[j] < i - 1 && a[i] - a[now[j] + 1] >= x) {
        now[j]++;
        s[j] = (s[j] + dp[now[j]][j]);
        if (s[j] > mod) s[j] -= mod;
      }
      dp[i][j + 1] = s[j];
    }
  }
  ret = 0;
  for (i = 1; i <= n; i++) {
    ret = (ret + dp[i][k]) % mod;
  }
  return ret;
}
void work() {
  long long i;
  scanf("%lld%lld", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= 100000 / (k - 1) + 5; i++) {
    tot[i] = solve(i);
  }
  for (i = 1; i <= 100000 / (k - 1) + 5; i++) {
    ans = (ans + (tot[i] - tot[i + 1] + mod) * i) % mod;
  }
  printf("%lld\n", ans);
}
int main() {
  work();
  return 0;
}
