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 INF = 1000 * 1000 * 1000 + 7;
const int _MOD = 998244353;
const long long INF64 = (long long)1e18;
const double EPS = 1e-8;
const int MN = 1 * 1000;
int dp[1001][1001];
int sum[1001];
int solve_for_x(vector<int>& A, int k, int x) {
  int n = A.size();
  int pointer = 0;
  dp[0][1] = 1;
  for (int i = 2; i <= k; ++i) {
    dp[0][i] = 0;
    sum[i - 1] = 0;
  }
  for (int i = 1; i < n; ++i) {
    dp[i][1] = 1;
    for (; A[pointer] + x <= A[i]; ++pointer) {
      for (int p = 1; p < k; p++) {
        sum[p] = ((sum[p] + dp[pointer][p]) % _MOD);
      }
    }
    for (int p = 2; p <= k; ++p) {
      dp[i][p] = sum[p - 1];
    }
  }
  int ANS = 0;
  for (int i = 0; i < n; ++i) {
    ANS = ((ANS + dp[i][k]) % _MOD);
  }
  return ANS;
}
int solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> A(n, 0);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
  }
  sort((A).begin(), (A).end());
  int ANS = 0;
  for (int x = int(A[n - 1] / (k - 1)) + 1; x >= 1; --x) {
    int buf = solve_for_x(A, k, x);
    ANS = ((ANS + buf) % _MOD);
  }
  printf("%d\n", ANS);
  return 0;
}
int main() {
  solve();
  return 0;
  freopen("in.1.txt", "r", stdin);
  int n_tests;
  scanf("%d", &n_tests);
  for (int i = 0; i < n_tests; i++) {
    clock_t begin = clock();
    solve();
    clock_t end = clock();
    printf("\n----- Time: %0.3f \n", (end - begin + 0.0) / CLOCKS_PER_SEC);
  }
  return 0;
}
