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;
void add(int &a, int b) {
  a += b;
  if (a >= 998244353) {
    a -= 998244353;
  }
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) {
    a += 998244353;
  }
}
int n, k, a[1000 << 1 | 5];
int dp[1000 << 1 | 5][1000 << 1 | 5];
int sum[1000 << 1 | 5];
int way[100005];
int calc(int d) {
  int ans = 0;
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = 0;
    }
  }
  dp[0][0] = 1;
  int p;
  for (int i = 0; i < k; i++) {
    sum[0] = dp[i][0];
    for (int j = 1; j <= n; j++) {
      sum[j] = (sum[j - 1] + dp[i][j]) % 998244353;
    }
    p = 1;
    for (int j = 1; j <= n; j++) {
      while (a[j] - a[p] >= d) {
        p++;
      }
      dp[i + 1][j] = sum[p - 1];
    }
  }
  for (int j = k; j <= n; j++) {
    add(ans, dp[k][j]);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  int dif = a[n] - a[1];
  int ans = 0;
  for (int i = 1; i * (k - 1) <= dif; i++) {
    way[i] = calc(i);
  }
  for (int i = 1; i * (k - 1) <= dif; i++) {
    sub(way[i], way[i + 1]);
    add(ans, (long long)i * way[i] % 998244353);
  }
  printf("%d\n", ans);
  return 0;
}
