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 li = long long;
using ld = long double;
void solve(bool);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  solve(true);
  return 0;
}
const int mod = 998244353;
int d[1003][1003];
int a[1003], n, k, nxt[1003];
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void solve(bool __attribute__((unused)) read) {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  li ans = 0;
  int delta = 0;
  while (delta <= a[n - 1] && clock() / (ld)CLOCKS_PER_SEC < 4.5) {
    ++delta;
    li cnt = 0;
    int j = 0;
    for (int i = 0; i < n; ++i) {
      memset(d[i], 0, sizeof(int) * (k + 1));
      while (j < n && a[j] - a[i] < delta) ++j;
      nxt[i] = j;
    }
    d[0][1] = 1;
    for (int i = 0; i < n; ++i) {
      if (i > 0) {
        for (int c = 1; c <= k; ++c) add(d[i][c], d[i - 1][c]);
      }
      for (int c = 1; c < k; ++c) {
        add(d[nxt[i]][c + 1], d[i][c]);
      }
      cnt += d[i][k];
    }
    ans = (ans + cnt) % mod;
  }
  cout << ans << "\n";
}
