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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxn = 510;
const int inf32 = 1e9;
const long long inf64 = 1e18;
const int mod = 998244353;
int mul(int first, int second) { return (long long)first * second % mod; }
int add(int first, int second) {
  long long ret = first;
  ret += second;
  if (ret >= mod) ret -= mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int N, K;
  cin >> N >> K;
  vector<int> a(N + 1);
  int up = 0;
  for (int i = 1; i <= N; ++i) cin >> a[i], up = max(up, a[i]);
  sort(a.begin() + 1, a.end());
  a[0] = 0;
  long long ans = 0;
  vector<int> pre(N + 1);
  vector<vector<int>> f(N + 1, vector<int>(K + 1, 0));
  vector<vector<int>> sum(N + 1, vector<int>(K + 1, 0));
  for (int first = 1; first <= up / (K - 1) + 1; ++first) {
    for (int i = 0; i <= N; ++i) {
      for (int j = 0; j <= K; ++j) {
        f[i][j] = sum[i][j] = 0;
      }
    }
    f[0][0] = sum[0][0] = 1;
    for (int i = 1; i <= N; ++i) sum[i][0] += sum[i - 1][0];
    int p = 0;
    for (int i = 1; i <= N; ++i) {
      while (p + 1 <= i && a[i] - a[p + 1] >= first) ++p;
      pre[i] = p;
    }
    for (int j = 1; j <= K; ++j) {
      for (int i = 1; i <= N; ++i) {
        f[i][j] = add(f[i][j], sum[pre[i]][j - 1]);
        sum[i][j] = add(sum[i][j], add(sum[i - 1][j], f[i][j]));
      }
    }
    long long v = 0;
    for (int i = 1; i <= N; ++i) {
      v = add(v, f[i][K]);
      ans = add(ans, f[i][K]);
    }
  }
  cout << ans << endl;
}
