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;
#pragma GCC optimize("O3")
const double esp = 1e-8;
const int mod = 998244353;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const int maxn = 1e3 + 5;
const int maxm = 1e6 + 5;
int a[maxn], n, k;
int dp[maxn][maxn], tsum[maxn][maxn];
int bound[maxn];
inline int add(int x, int y) {
  return (x + y) >= mod ? (x + y - mod) : (x + y);
}
int solve(int x) {
  bound[0] = 0;
  for (int i = 1; i <= n; i++) {
    bound[i] = bound[i - 1];
    while (a[i] - a[bound[i] + 1] >= x) bound[i]++;
  }
  tsum[1][0] = 0;
  for (int i = 1; i <= k; i++) {
    for (int j = i; j <= n; j++) {
      if (i == 1) {
        dp[i][j] = 1;
        tsum[i][j] = (tsum[i][j - 1] + dp[i][j]);
        continue;
      }
      dp[i][j] = tsum[i - 1][bound[j]];
      tsum[i][j] = add(tsum[i][j - 1], dp[i][j]);
    }
  }
  return tsum[k][n];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int ans = 0;
  int maxt = (a[n] - a[1]) / (k - 1);
  for (int t = 1; t <= maxt; t++) {
    ans = add(ans, solve(t));
  }
  cout << ans << endl;
  return 0;
}
