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 n;
int k;
vector<int> v;
int dp[1003][1003];
int im[1003][1003];
int solve(int want) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = 0;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      im[i][j] = 0;
    }
  }
  for (int i = 0; i < n; i++) {
    dp[i][1] = 1;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int z = lower_bound(v.begin() + i + 1, v.end(), v[i] + want) - v.begin();
    for (int j = 0; j <= k; j++) {
      dp[i][j] += im[i][j];
      im[i + 1][j] += im[i][j];
      if (dp[i][j] >= 998244353) dp[i][j] -= 998244353;
      if (im[i + 1][j] >= 998244353) im[i + 1][j] -= 998244353;
      int go = v[i] + want;
      im[z][j + 1] += dp[i][j];
      if (im[z][j + 1] >= 998244353) im[z][j + 1] -= 998244353;
    }
    ans += dp[i][k];
    if (ans >= 998244353) ans -= 998244353;
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    v.push_back(a);
  }
  sort(v.begin(), v.end());
  int num_interval = k - 1;
  int maxi = v.back() / num_interval;
  long long int ans = 0;
  for (int j = 1; j <= maxi; j++) {
    ans += solve(j);
  }
  ans %= 998244353;
  printf("%lld\n", ans);
  return 0;
}
