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>
const int MOD = 998244353;
int n, k;
std::vector<int> A;
int binco[1005][1005];
int dp[1005][1005], dp2[1005][1005];
int get(int r) {
  if ((k - 1) * r > 100005) return 0;
  for (int i = 0; i < (n + 1); ++i) {
    for (int j = 0; j < (k + 1); ++j) {
      dp[i][j] = 0;
      dp2[i][j] = 0;
    }
  }
  std::vector<int> next(n);
  int right = 0;
  for (int i = 0; i < (n); ++i) {
    while (right < n && A[right] - A[i] < r) right++;
    next[i] = right;
  }
  int ans = 0;
  dp[n][0] = 1;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = (0); j < (k + 1); ++j) {
      dp[i][j] = ((j == 0 ? 0 : dp[next[i]][j - 1]) + dp[i + 1][j]) % MOD;
    }
  }
  return dp[0][k];
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cin >> n >> k;
  A.resize(n);
  for (int i = 0; i < (n); ++i) std::cin >> A[i];
  std::sort(A.begin(), A.end());
  int ans = 0;
  for (int i = (1); i < (100002); ++i) ans = (ans + get(i)) % MOD;
  std::cout << ans;
  return 0;
}
