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 n, a[N], k, dp[N][N], ans;
inline void addmod(int &a, int b) { a = (a + b) % mod; }
inline void mulmod(int &a, int b) { a = 1ll * a * b % mod; }
inline int add(int a, int b) { return (a + b) % mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1);
  int lst = 0;
  dp[0][0] = 1, a[0] = -1e9;
  for (int b = 1; b <= (a[n] - a[1]) / (k - 1) + 1; ++b) {
    for (int len = 1; len <= k; ++len) {
      int s = 0, l = 0;
      for (int i = 1; i <= n; ++i) {
        while (l < i && a[i] - a[l] >= b) addmod(s, dp[len - 1][l++]);
        dp[len][i] = s;
      }
    }
    int sum = 0;
    for (int i = 1; i <= n; ++i) addmod(sum, dp[k][i]);
    addmod(ans, mul((lst - sum + mod) % mod, b - 1));
    lst = sum;
  }
  cout << ans;
}
