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;
inline int lgput(int a, int b) {
  int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (1LL * ans * a) % MOD;
    b >>= 1;
    a = (1LL * a * a) % MOD;
  }
  return ans;
}
inline void mod(int &x) {
  if (x >= MOD) x -= MOD;
}
inline void add(int &x, int y) {
  x += y;
  mod(x);
}
inline void sub(int &x, int y) {
  x += MOD - y;
  mod(x);
}
inline void mul(int &x, int y) { x = (1LL * x * y) % MOD; }
inline int inv(int x) { return lgput(x, MOD - 2); }
using namespace std;
int main() {
  int i, j, n, k;
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> k;
  vector<int> a(n + 1);
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<vector<int> > dp(k + 1, vector<int>(n + 1));
  vector<vector<int> > sp = dp;
  int last = 0, ans = 0;
  for (int val = 1e5 / (k - 1); val > 0; val--) {
    for (i = 1; i <= n; i++) {
      sp[1][i] = i;
    }
    for (j = 2; j <= k; j++) {
      int pos = 1;
      for (i = j; i <= n; i++) {
        while (a[i] - a[pos] >= val) {
          pos++;
        }
        dp[j][i] = sp[j - 1][pos - 1];
        sp[j][i] = dp[j][i];
        add(sp[j][i], sp[j][i - 1]);
      }
    }
    ans = (ans + 1LL * (sp[k][n] - last + MOD) * val) % MOD;
    last = sp[k][n];
  }
  cout << ans;
  return 0;
}
