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 = 1e3 + 10;
const long long mod = 998244353;
int n, k;
int a[MAXN];
long long dp[MAXN][MAXN];
set<int> diffs;
long long cal(int lb) {
  dp[0][1] = 1;
  for (int j = 2; j <= k; j++) dp[0][j] = 0;
  int pre = -1;
  for (int i = 1; i < n; i++) {
    while (a[pre + 1] <= a[i] - lb && pre + 1 < i) pre++;
    dp[i][1] = i + 1;
    for (int j = 2; j <= k; j++) {
      dp[i][j] = dp[i - 1][j];
      if (pre >= 0) dp[i][j] += dp[pre][j - 1];
      dp[i][j] %= mod;
    }
  }
  return dp[n - 1][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (a[i] != a[j]) diffs.insert(a[j] - a[i]);
    }
  }
  long long res = 0;
  int pre = 0;
  for (int diff : diffs) {
    if (diff * (k - 1) > a[n - 1] - a[0]) break;
    res += (diff - pre) * cal(diff) % mod;
    res %= mod;
    pre = diff;
  }
  cout << res << endl;
  return 0;
}
