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>
#pragma GCC optimize("Ofast")
using namespace std;
const int MOD = 998244353;
int n, k;
int ans;
int v[1005], dp[1005][1005], sp[1005][1005];
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int solve(int val) {
  int l = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][1] = 1;
    sp[i][1] = sp[i - 1][1] + 1;
    while (v[l] + val <= v[i]) l++;
    if (v[l] + val > v[i]) l = max(l - 1, 0);
    for (int j = 2; j <= min(i, k); j++) {
      dp[i][j] = sp[l][j - 1];
      sp[i][j] = sp[i - 1][j];
      add(sp[i][j], dp[i][j]);
    }
  }
  int ans = 0;
  for (int i = k; i <= n; i++) add(ans, dp[i][k]);
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> v[i];
  sort(v + 1, v + n + 1);
  for (int val = 1; val <= v[n] / (k - 1); val++)
    ans = (ans + solve(val)) % MOD;
  cout << ans;
  return 0;
}
