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;
const int N = 1e3 + 7;
const int mod = 998244353;
int a[N];
int dp[N][N];
int n, k;
int solve(int lim) {
  for (int j = 2; j <= k; ++j) {
    int sum = 0, nowi = j - 1;
    for (int i = j; i <= n; ++i) {
      while (a[nowi] + lim <= a[i] && nowi < i) {
        sum = sum + dp[nowi][j - 1];
        if (sum >= mod) sum -= mod;
        nowi++;
      }
      dp[i][j] = sum;
    }
  }
  int ret = 0;
  for (int i = k; i <= n; ++i) {
    ret = ret + dp[i][k];
    if (ret >= mod) ret -= mod;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) {
    dp[i][1] = 1;
  }
  int ans = 0;
  for (int i = 1; i * (k - 1) <= a[n] - a[1]; ++i) {
    ans = ans + solve(i);
    if (ans >= mod) ans -= mod;
  }
  cout << ans;
}
