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 = 1100;
const int M = 260;
const long long mod = 998244353;
const long long MOD = 998244353;
const long double eps = 0.000000001;
const int P = 1336;
const int inf = 1e16;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
int n, k;
vector<pair<int, int> > a;
int dp[N][N];
long long solve(int x) {
  for (int j = 0; j < k; j++) {
    for (int i = 0; i < n; i++) {
      dp[j][i] = 0;
      if (j == 0) dp[j][i] = 1;
    }
  }
  long long ans = 0;
  for (int j = 1; j < k; j++) {
    int uk = 0;
    long long s = 0;
    for (int i = 0; i < n; i++) {
      while (uk < n && a[uk].first + x <= a[i].first) {
        s += dp[j - 1][uk];
        s %= mod;
        uk++;
      }
      dp[j][i] = s;
      if (j == k - 1) {
        ans += dp[j][i];
        ans %= mod;
      }
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  cin >> n >> k;
  a.resize(n);
  int ma = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    ma = max(ma, a[i].first);
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  long long ans = 0, s = 0;
  for (int i = ma / (k - 1); i >= 1; i--) {
    long long c = solve(i);
    c -= s;
    c += mod;
    c %= mod;
    s += c;
    s %= mod;
    c *= i;
    c %= mod;
    ans += c;
    ans %= mod;
  }
  cout << ans;
}
