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 n, k;
int a[1007];
long long ans = 0;
long long dp[2][1007];
long long pref[1007];
long long prv_ans;
long long calc(int sr) {
  int curr = 0, prv = 1;
  for (int i = 1; i <= n; i++) {
    dp[prv][i] = 1;
  }
  for (int j = 2; j <= k; ++j) {
    for (int i = 1; i <= n; ++i) {
      pref[i] = pref[i - 1] + dp[prv][i];
      pref[i] = (pref[i] >= 998244353) ? pref[i] - 998244353 : pref[i];
    }
    int init = 0;
    for (int i = 1; i <= n; ++i) {
      while (init < n && a[i] - sr >= a[init + 1]) {
        ++init;
      }
      dp[curr][i] = pref[init];
    }
    prv ^= 1;
    curr ^= 1;
  }
  long long sm = 0;
  for (int i = 1; i <= n; ++i) {
    sm = (sm + dp[prv][i]);
    sm = (sm >= 998244353) ? sm - 998244353 : sm;
  }
  long long ret = (sm - prv_ans);
  ret = (ret >= 0) ? ret : ret + 998244353;
  prv_ans = sm;
  return ret;
}
void input() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
}
void solve() {
  int mxval = (a[n] - a[1] + k - 2) / (k - 1);
  for (int i = mxval; i >= 1; --i) {
    ans = (ans + (calc(i) * i) % 998244353);
    ans = (ans >= 998244353) ? ans - 998244353 : ans;
  }
  printf("%I64d\n", ans);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
