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 MAXN = 1050;
const int mod = 998244353;
int n, k;
int a[MAXN];
long long ans = 0;
long long dp[MAXN][MAXN];
long long sum[MAXN][MAXN];
long long getAns(int x) {
  if (x > 0) {
    int maxK = a[n - 1] / x + 1;
    if (maxK < k) {
      return 0;
    }
  }
  int p = 0;
  for (int i = 0; i < n; i++) {
    while (p + 1 < i && a[i] - a[p + 1] >= x) p++;
    for (int j = 1; j <= k; j++) {
      if (j == 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] = 0;
        if (p < i && a[i] - a[p] >= x) {
          dp[i][j] = (dp[i][j] + sum[p][j - 1]) % mod;
        }
      }
      sum[i][j] = dp[i][j];
      if (i > 0) sum[i][j] = (sum[i][j] + sum[i - 1][j]) % mod;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; i++) res = (res + dp[i][k]) % mod;
  return res;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  long long prevAns = 0;
  for (int i = 0; i <= a[n - 1] + 1; i++) {
    long long curAns = getAns(i);
    if (i > 0) {
      long long diff = (prevAns - curAns + mod) % mod;
      long long add = diff * (i - 1) % mod;
      ans = (ans + add) % mod;
    }
    prevAns = curAns;
  }
  cout << ans << endl;
  return 0;
}
