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;
const int N = 1e3 + 10;
int power(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  int n, k, a[N], min_id[N], dp[N][N], ans = 0;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    min_id[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= a[n] - a[1]; i++) {
    dp[0][0] = 1;
    for (int j = 1; j <= k && j * i <= a[n] - a[1]; j++) {
      dp[0][j] = 0;
    }
    for (int j = 1; j <= n; j++) {
      while (min_id[j] > 0 && a[j] - a[min_id[j]] < i) min_id[j]--;
      dp[j][0] = 1;
      for (int cnt = 1; cnt <= k && (cnt - 1) * i <= a[n] - a[1]; cnt++) {
        dp[j][cnt] = (dp[min_id[j]][cnt - 1] + dp[j - 1][cnt]) % mod;
      }
    }
    ans += (1LL * (k - 1) * i) <= a[n] - a[1] ? dp[n][k] : 0;
    ans %= mod;
  }
  printf("%d", ans);
  return 0;
}
