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 ll = long long;
const int MOD = 998244353;
void selfAdd(int &a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
}
void selfSub(int &a, int b) {
  a -= b;
  if (a < 0) {
    a += MOD;
  }
}
int add(int a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int mult(int a, int b) { return (ll)a * b % MOD; }
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  int a[1 + N];
  for (int i = 1; i <= N; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + N + 1);
  int B = 100001 / (K - 1) + 1;
  int SumCount[1 + N][2];
  int beauty[1 + B + 1];
  memset(beauty, 0, sizeof(beauty));
  for (int b = 1; b <= B; b++) {
    for (int n = 0; n <= N; n++) {
      SumCount[n][0] = 1;
    }
    for (int k = 1; k <= K; k++) {
      for (int n = 0; n <= N; n++) {
        SumCount[n][k % 2] = 0;
      }
      int n = 1;
      int oldN = 1;
      while (n <= N) {
        if (a[n] - a[oldN] < b) {
          SumCount[n][k % 2] =
              add(SumCount[n - 1][k % 2], SumCount[oldN - 1][(k - 1) % 2]);
          n++;
        } else {
          oldN++;
        }
      }
    }
    beauty[b] = SumCount[N][K % 2];
  }
  int answer = 0;
  for (int b = 1; b <= B; b++) {
    selfSub(beauty[b], beauty[b + 1]);
    selfAdd(answer, mult(b, beauty[b]));
  }
  printf("%d\n", answer);
  return 0;
}
