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 const N = 1000, M = 100000, md = 998244353;
int x[N + 1], dp[N + 1][2], pr[N + 1][2];
inline void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < (int)(n + 1); ++i) scanf("%d", x + i);
  sort(x + 1, x + n + 1);
  int an = 0;
  for (int i = 1; i < (int)(M / (k - 1) + 1); ++i) {
    for (int j = 0; j < (int)(n + 1); ++j)
      for (int w = 0; w < (int)(2); ++w) dp[j][w] = 0;
    dp[0][0] = 1;
    for (int j = 0; j < (int)(k); ++j) {
      int a = 0, b = 0;
      pr[0][0] = j == 0;
      for (int w = 1; w < (int)(n + 1); ++w) {
        for (int z = 0; z < (int)(2); ++z)
          ad(pr[w][z] = pr[w - 1][z], dp[w][z]);
        while (a < w && x[w] - x[a + 1] > i) ++a;
        while (b < w && x[w] - x[b + 1] >= i) ++b;
        for (int z = 0; z < (int)(2); ++z) dp[w][z] = 0;
        for (int z = 0; z < (int)(2); ++z) ad(dp[w][z], pr[a][z]);
        if (a != b)
          for (int z = 0; z < (int)(2); ++z) {
            ad(dp[w][1], pr[b][z]);
            ad(dp[w][1], md - pr[a][z]);
          }
      }
    }
    for (int j = 2; j < (int)(n + 1); ++j) ad(an, (long long)dp[j][1] * i % md);
  }
  printf("%d\n", an);
}
