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 = 2e5 + 5;
const int MOD = 998244353;
const int M = 1005;
int lower[N];
int arr[M];
int dp[M][M], vis[M][M];
int n, k, visid, x;
bool flag;
int solve(int idx, int take) {
  if (take == k) return flag = 1;
  if (idx == n) return 0;
  int &ret = dp[idx][take];
  if (vis[idx][take] == visid) return ret;
  vis[idx][take] = visid;
  ret = solve(idx + 1, take) + solve(lower[arr[idx] + x], take + 1);
  if (ret >= MOD) ret -= MOD;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(lower, -1, sizeof lower);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  for (int i = n - 1; ~i; i--) {
    lower[arr[i]] = i;
  }
  lower[N - 1] = n;
  for (int i = N - 1; ~i; i--) {
    if (~lower[i]) continue;
    lower[i] = lower[i + 1];
  }
  int ans = 0;
  flag = true;
  for (int i = 1; flag; i++) {
    flag = false;
    visid++;
    x = i;
    ans += solve(0, 0);
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans << endl;
  return 0;
}
