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;
int a[1005], N, K, M = 0;
int DP[1005][1005];
int sum[1005][1005];
long long solve(int x) {
  int i, j, k = 0;
  for (i = 0; i <= N; ++i) {
    for (j = 0; j <= K; ++j) {
      DP[i][j] = sum[i][j] = 0;
    }
  }
  for (i = 1; i <= N; ++i) DP[i][1] = 1;
  sum[1][1] = 1;
  for (i = 2; i <= N; ++i) {
    while (k + 1 < i && a[k + 1] - a[i] >= x) ++k;
    for (j = 2; j <= K; ++j) {
      DP[i][j] = sum[k][j - 1];
      if (DP[i][j] >= 998244353) DP[i][j] -= 998244353;
    }
    for (j = 0; j <= K; ++j) {
      sum[i][j] = DP[i][j] + sum[i - 1][j];
      if (sum[i][j] >= 998244353) sum[i][j] -= 998244353;
    }
  }
  int ret = 0;
  for (i = 1; i <= N; ++i) {
    ret += DP[i][K];
    if (ret >= 998244353) ret -= 998244353;
  }
  return ret;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &a[i]);
    M = max(M, a[i]);
  }
  a[0] = M + 1;
  sort(a + 1, a + 1 + N);
  reverse(a + 1, a + 1 + N);
  int ret = 0;
  for (int x = 1; x <= M / (K - 1); ++x) {
    ret += solve(x);
    if (ret >= 998244353) ret -= 998244353;
  }
  printf("%d\n", ret);
}
