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 mod = 998244353;
int dp[1234][1234], last[1234], sum[1234], n, k;
int arr[1234];
int solve(int val) {
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  dp[0][0] = 1;
  last[0] = 0;
  for (int i = 1; i <= n; i++) {
    last[i] = last[i - 1];
    while (arr[i] - arr[last[i] + 1] >= val) last[i]++;
  }
  for (int i = 0; i < k; i++) {
    sum[0] = dp[i][0];
    for (int j = 1; j <= n; j++) sum[j] = (sum[j - 1] + dp[i][j]) % mod;
    for (int j = 1; j <= n; j++) dp[i + 1][j] = sum[last[j]];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += dp[k][i], ans %= mod;
  return ans;
}
int main() {
  ios::sync_with_stdio();
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  sort(arr + 1, arr + n + 1);
  int mx = arr[n], ans = 0;
  for (int i = 1; (k - 1) * i <= arr[n]; i++) ans = (ans + solve(i)) % mod;
  cout << ans << endl;
}
