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;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int maxn = 1e3 + 100;
int n, k;
long long a[maxn];
long long dp[maxn][maxn];
long long sum(long long x, long long y) { return (x + y) % mod; }
long long gao(int x) {
  dp[0][0] = 1;
  int now = 0;
  for (int i = 1; i < n + 1; i++) {
    while (a[i] - a[now + 1] >= x) now++;
    dp[i][0] = dp[i - 1][0];
    for (int j = 1; j < k + 1; j++) {
      dp[i][j] = (dp[i - 1][j] + dp[now][j - 1]) % mod;
    }
  }
  return dp[n][k];
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (int i = 1; i * (k - 1) <= a[n]; i++) {
    ans = sum(ans, gao(i));
  }
  cout << ans << endl;
  return 0;
}
