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 double inf = 2e18;
const int maxn = 100020;
const long long mod = 998244353;
map<long long, long long> num;
int n, k, a[maxn];
int f[1020][1020], sum[1020][1020], ans;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int x = 1; x <= a[n] / (k - 1); x++) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= k; j++) sum[i][j] = f[i][j] = 0;
    for (int i = 1; i <= n; i++) sum[i][1] = i;
    for (int j = 2; j <= k; j++) {
      int id = 0;
      for (int i = 2; i <= n; i++) {
        while (id < i && a[i] - a[id + 1] >= x) id++;
        f[i][j] = sum[id][j - 1];
      }
      for (int i = 1; i <= n; i++) sum[i][j] = (f[i][j] + sum[i - 1][j]) % mod;
    }
    for (int i = 1; i <= n; i++) ans = (ans + f[i][k]) % mod;
  }
  cout << ans << endl;
}
