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;
constexpr int MOD = 998244353;
constexpr int MAXN = 1e5 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
ll A[MAXN];
ll memos[MAXN];
ll pre_sum[MAXN];
int pre_idx[MAXN];
ll Solution() {
  sort(A, A + N);
  memset(pre_idx, 0, sizeof(pre_idx));
  for (int i = (int)(0); i < (int)(N); ++i) pre_idx[A[i]]++;
  for (int i = (int)(1); i < (int)(MAXN); ++i) pre_idx[i] += pre_idx[i - 1];
  ll maxi = A[N - 1];
  ll ans = 0;
  for (int p = (int)(1); p < (int)(maxi / (K - 1) + 1); ++p) {
    pre_sum[0] = 1;
    for (int i = (int)(1); i < (int)(N); ++i) {
      pre_sum[i] = pre_sum[i - 1] + 1;
    }
    for (int i = (int)(2); i < (int)(K + 1); ++i) {
      for (int j = (int)(0); j < (int)(N); ++j) {
        memos[j] = 0;
        int g = A[j] - p >= 0 ? pre_idx[A[j] - p] : 0;
        if (g != 0) {
          memos[j] = pre_sum[g - 1];
        }
      }
      pre_sum[0] = memos[0];
      for (int j = (int)(1); j < (int)(N); ++j) {
        pre_sum[j] = (pre_sum[j - 1] + memos[j]) % MOD;
      }
    }
    ans += pre_sum[N - 1];
    ans %= MOD;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N >> K) {
    for (int i = (int)(0); i < (int)(N); ++i) cin >> A[i];
    cout << Solution() << '\n';
  }
  return 0;
}
