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 int MOD = 998244353;
int n, k;
int a[1000];
int cnt[1 << 20];
int dp[1000][1000 + 2];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  int max = (a[n - 1] - a[0]) / (k - 1);
  int ans = 0;
  for (int x = 1; x <= max; x++) {
    int sum[1000 + 2];
    sum[0] = 1;
    for (int j = 1; j <= k; j++) {
      sum[j] = 0;
    }
    int cur = -1;
    for (int i = 0; i < n; i++) {
      while (cur + 1 < i && a[i] - a[cur + 1] >= x) {
        cur++;
        for (int j = 1; j <= k; j++) {
          sum[j] += dp[cur][j];
          sum[j] %= MOD;
        }
      }
      for (int j = 1; j <= k; j++) {
        dp[i][j] = sum[j - 1];
      }
    }
    for (int i = 0; i < n; i++) {
      cnt[x] += dp[i][k];
      cnt[x] %= MOD;
    }
    ans += cnt[x];
    ans %= MOD;
  }
  cout << ans << endl;
}
