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;
using INT = long long;
const int NN = 1010;
const int MM = 101010;
const int mod = 998244353;
int a[NN], pos[MM], dp[NN][NN];
int add(int &u, int v) {
  u += v;
  if (u >= mod) u -= mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i < MM; i++) {
    pos[i] = lower_bound(a + 1, a + n + 1, i) - a;
  }
  for (int i = MM / (k - 1), pre = 0; i; i--) {
    for (int j = n; j; j--) {
      dp[j][1] = (n - j + 1);
      for (int t = 2; t <= k; t++) {
        int s = a[j] + i;
        dp[j][t] = 0;
        if (s < MM and pos[s] <= n) {
          dp[j][t] = dp[pos[s]][t - 1];
        }
        dp[j][t] = (dp[j][t] + dp[j + 1][t]) % mod;
      }
    }
    ans = (ans + (INT)i * (dp[1][k] - pre + mod) % mod) % mod;
    pre = dp[1][k];
  }
  cout << ans << endl;
}
