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 N = 1005;
long long dp[N][N], tot[N][N], ans = 0, ret[N * N];
int n, a[N], k;
void solve(int x) {
  dp[0][0] = 1;
  tot[0][0] = 1;
  int now = 0;
  for (int i = 1; i <= n; ++i) {
    while ((a[i] - a[now]) >= x) ++now;
    for (int j = 0; j <= k; ++j) {
      if (j > 0) dp[i][j] = tot[now - 1][j - 1];
      tot[i][j] = (tot[i - 1][j] + dp[i][j]) % mod;
    }
    ret[x] = (ret[x] + dp[i][k]) % mod;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  a[0] = -1000000000;
  for (int i = 1; i * (k - 1) <= a[n]; ++i) {
    solve(i);
    ans = (ans + ret[i]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
