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 long long mod = 998244353;
const int N = 1e3 + 5;
long long pwr(long long x, long long y);
int a[N], n, k;
int dp[N][N];
int solve(int x) {
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < k + 1; j++) dp[i][j] = 0;
  }
  dp[0][0] = 1;
  int pos = 0;
  for (int i = 1; i < n + 1; i++) {
    while ((a[i] - a[pos + 1]) >= x) ++pos;
    for (int j = 0; j < k + 1; j++) {
      dp[i][j] = dp[i - 1][j] % mod;
    }
    for (int j = 1; j < k + 1; j++) {
      dp[i][j] = (dp[i][j] + dp[pos][j - 1]) % mod;
    }
  }
  return dp[n][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i < 1e5 / (k - 1) + 1; i++) {
    ans = (ans + solve(i)) % mod;
  }
  cout << ans << endl;
  return 0;
}
long long pwr(long long x, long long y) {
  long long ans = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) ans = (x * (ans % mod)) % mod;
    x = (x * x) % mod;
    y = y / 2;
  }
  return ans % mod;
}
