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 maxn = 1005;
const int MAXM = 1e5 + 7;
const int mod = 998244353;
int arr[maxn];
long long dp[maxn][maxn];
int n, k;
long long solve(int x) {
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  for (int i = 0; i <= n; i++) dp[1][i] = i;
  for (int i = 1; i < k; i++) {
    int p = 0;
    for (int j = 0; j < n; j++) {
      while (arr[j] - arr[p] >= x) p++;
      dp[i + 1][j + 1] = (dp[i + 1][j] + dp[i][p]) % mod;
    }
  }
  return dp[k][n];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  sort(arr, arr + n);
  long long ans = 0;
  for (int i = 1; i * (k - 1) <= MAXM; i++) ans = (ans + solve(i)) % mod;
  printf("%lld\n", ans);
}
