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 = 1007;
int a[maxn];
long long dp[maxn][maxn], sum[maxn][maxn], res[maxn * 100];
vector<int> v;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) v.push_back(a[j] - a[i]);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  int sz = v.size();
  for (int ind = 0; ind < sz; ind++) {
    int d = v[ind];
    if (d > a[n] / (k - 1) + 1) break;
    int pos = 0;
    for (int i = 1; i <= n; i++) {
      dp[i][1] = 1;
      sum[i][1] = ((sum[i - 1][1] + dp[i][1]) % 998244353);
      while (a[i] - a[pos + 1] >= d && pos + 1 < i) pos++;
      for (int len = 2; len <= k; len++) {
        dp[i][len] = sum[pos][len - 1];
        sum[i][len] = ((sum[i - 1][len] + dp[i][len]) % 998244353);
      }
    }
    res[ind] = sum[n][k];
  }
  long long ans = 0;
  for (int ind = 0; ind < sz; ind++) {
    int d = v[ind];
    if (d > a[n] / (k - 1) + 1) break;
    ans = ((ans + ((((res[ind] - res[ind + 1] + 998244353) % 998244353) *
                    (long long)d) %
                   998244353)) %
           998244353);
  }
  printf("%lld\n", ans);
  return 0;
}
