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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = (int)998244353;
const int N = (int)1e3 + 7;
int n, k, a[N], dp[N][N];
inline int add(int a, int b) { return (a += b) >= MOD ? a - MOD : a; }
int work(int d) {
  for (int i = (0); i < (n + 1); ++i)
    for (int j = (0); j < (k + 1); ++j) dp[i][j] = 0;
  int pre = 0;
  for (int i = (0); i < (n + 1); ++i) dp[i][0] = 1;
  for (int i = (1); i < (n + 1); ++i) {
    while (a[pre + 1] + d <= a[i]) pre++;
    for (int j = (1); j < (k + 1); ++j) {
      dp[i][j] = add(dp[i - 1][j], dp[pre][j - 1]);
    }
  }
  return dp[n][k];
}
void Init() {
  for (int i = (1); i < (n + 1); ++i) scanf("%d", a + i);
}
int Solve() {
  sort(a + 1, a + n + 1);
  int up = a[n] / (k - 1);
  int ans = 0;
  for (int i = (1); i < (up + 1); ++i) {
    ans = add(ans, work(i));
  }
  return printf("%d\n", ans);
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    Init();
    Solve();
  }
  return 0;
}
