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 = 1100;
const long long MOD = 998244353;
int N, K;
int arr[MAXN];
int dp[MAXN][MAXN];
int ps[MAXN][MAXN];
long long figure(int k) {
  if (k * (K - 1) > arr[N - 1]) return 0;
  for (int i = 0; i <= N; i++)
    for (int j = 0; j <= K; j++) dp[i][j] = ps[i][j] = 0;
  for (int i = 1; i <= N; i++) dp[i][1] = 1;
  int cstop = 0;
  for (int i = 0; i < N; i++) {
    while (cstop < i && arr[i] - arr[cstop] >= k) cstop++;
    for (int j = 1; j < K; j++) {
      dp[i + 1][j + 1] = ps[cstop][j];
    }
    for (int j = 0; j <= K; j++) {
      ps[i + 1][j] = (ps[i][j] + dp[i + 1][j]) % MOD;
    }
  }
  return ps[N][K];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> arr[i];
  sort(arr, arr + N);
  long long ans = 0;
  for (int i = 1; i <= 100000; i++) ans = (ans + figure(i)) % MOD;
  cout << ans << "\n";
}
