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 = 1005, Mod = 998244353;
int a[N], dp[N][N], ans[100005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -1e9;
  int maxv = a[n] / (k - 1);
  for (int v = 0; v <= maxv; v++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= k; j++) dp[i][j] = 0;
    for (int i = 1; i <= n; i++) dp[i][1] = 1;
    for (int j = 2; j <= k; j++) {
      int last = 1, sum = 0;
      for (int i = 2; i <= n; i++) {
        while (a[i] - a[last] >= v && last < i)
          sum = (sum + dp[last][j - 1]) % Mod, ++last;
        dp[i][j] = (dp[i][j] + sum) % Mod;
      }
    }
    for (int i = k; i <= n; i++) ans[v] = (ans[v] + dp[i][k]) % Mod;
  }
  int Ans = 0;
  for (int i = 1; i <= maxv; i++) ans[i] = (ans[i] - ans[i + 1] + Mod) % Mod;
  for (int i = 1; i <= maxv; i++) Ans = (Ans + 1ll * ans[i] * i % Mod) % Mod;
  printf("%d", Ans);
  return 0;
}
