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 long long mod = 998244353;
const int maxn = 1e3 + 5;
const int N = 1e5 + 5;
int n, k, tot, maxx;
map<int, int> mp;
pair<int, int> a[maxn];
long long dp[maxn][maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    mp[x]++;
    maxx = max(maxx, x);
  }
  for (auto it : mp) a[++tot] = it;
  long long ans = 0, last = 0;
  for (int c = maxx; c >= 1; c--) {
    if (a[1].first + 1ll * c * (k - 1) > a[tot].first) continue;
    for (int i = 0; i < maxn; i++) dp[i][0] = 1;
    int pos = 0;
    for (int i = 1; i <= tot; i++) {
      for (int j = 1; j <= k; j++) dp[i][j] = dp[i - 1][j];
      while (a[pos + 1].first + c <= a[i].first) ++pos;
      for (int j = 1; j <= k; j++) {
        dp[i][j] = (dp[i][j] + dp[pos][j - 1] * a[i].second % mod) % mod;
      }
    }
    ans = (ans + (dp[tot][k] - last) * (c)) % mod;
    ans = (ans + mod) % mod;
    last = dp[tot][k];
  }
  cout << ans << '\n';
  return 0;
}
