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 = 1010;
const int W = 1e5 + 10;
const long long MOD = 998244353;
int n, k;
int a[N];
long long dp[N][N], qzh[N][N];
int main() {
  scanf("%d%d", &n, &k);
  int maxs = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    maxs = max(maxs, a[i]);
  }
  sort(a + 1, a + 1 + n);
  long long be = 0, ans = 0;
  for (long long lim = 0; lim <= maxs / (k - 1) + 1; ++lim) {
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= k; ++j) {
        dp[i][j] = qzh[i][j] = 0;
      }
    }
    dp[0][0] = qzh[0][0] = 1;
    int l = 1;
    long long now = 0;
    for (int i = 1; i <= n; ++i) {
      while (l + 1 <= i && a[i] - a[l] >= lim) ++l;
      for (int j = 0; j <= k; ++j) {
        if (j == 0)
          dp[i][j] = 0;
        else {
          dp[i][j] = qzh[j - 1][l - 1];
        }
        qzh[j][i] = (qzh[j][i - 1] + dp[i][j]) % MOD;
      }
      now += dp[i][k];
    }
    if (lim != 0) {
      (ans += (lim - 1) * ((be - now) % MOD) % MOD + MOD) %= MOD;
    }
    be = now;
  }
  printf("%lld\n", ans);
}
