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 MAXN = 1010;
const int MAXV = 1E5 + 10;
const int P = 998244353;
template <typename T1, typename T2>
void Inc(T1 &a, T2 b) {
  a += b;
  if (a >= P) a -= P;
}
template <typename T1, typename T2>
void Dec(T1 &a, T2 b) {
  a -= b;
  if (a < 0) a += P;
}
template <typename T1, typename T2>
T1 Add(T1 a, T2 b) {
  a += b;
  return a >= P ? a - P : a;
}
template <typename T1, typename T2>
T1 Sub(T1 a, T2 b) {
  a -= b;
  return a < 0 ? a + P : a;
}
int n, k, a[MAXN];
int f[MAXN][MAXN], sum[MAXN];
long long Ans;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  int lst = 0;
  for (int s = a[n] / (k - 1); ~s; s--) {
    for (int i = 1; i <= k; i++) sum[i] = 0;
    int ptr = 0, tot = 0;
    for (int i = 1; i <= n; i++) {
      while (ptr + 1 <= i && a[i] - a[ptr + 1] >= s) {
        ++ptr;
        for (int j = 1; j <= k; j++) Inc(sum[j], f[ptr][j]);
      }
      for (int j = 1; j <= k; j++) f[i][j] = sum[j - 1];
      Inc(f[i][1], 1);
      Inc(tot, f[i][k]);
    }
    Inc(Ans, 1ll * Sub(tot, lst) * s % P);
    lst = tot;
  }
  printf("%lld\n", Ans);
}
