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 mx = 1e3 + 10;
const int mod = 998244353;
int n, k;
int a[mx];
long long dp[mx][mx];
long long qz[mx][mx];
int main() {
  long long ans = 0;
  cin >> n >> k;
  int maxa = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  a[0] = -0x3f3f3f3f;
  maxa = a[n] / (k - 1);
  for (int j = 1; j <= n; ++j) {
    qz[1][j] = qz[1][j - 1] + 1;
  }
  for (int i = 1; i <= maxa; ++i) {
    for (int j = 2; j <= k; ++j) {
      int pos = 0;
      for (int x = 1; x <= n; ++x) {
        while (a[x] - a[pos] >= i) ++pos;
        --pos;
        dp[j][x] = qz[j - 1][pos];
        qz[j][x] = (qz[j][x - 1] + dp[j][x]) % mod;
      }
      if (j == k) ans = (ans + qz[j][n]) % mod;
    }
  }
  cout << ans << endl;
}
