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>
#pragma GCC optimize(2)
using namespace std;
const int N = 1e3 + 10;
const long long md = 998244353;
int n, a[N], k;
long long f[N][N], ans, dp[100010];
long long sol(int x) {
  int i, j;
  for (i = 1; i <= n; ++i) f[i][1] = (long long)i;
  for (j = 2; j <= k; ++j) {
    for (i = 1; i <= n; ++i) {
      if (a[i] - a[1] < x) {
        f[i][j] = 0;
        continue;
      }
      int l = 1, r = i - 1, p;
      while (l < r - 1) {
        int mid = l + r >> 1;
        if (a[i] - a[mid] >= x)
          l = mid;
        else
          r = mid - 1;
      }
      if (a[i] - a[r] >= x)
        p = r;
      else
        p = l;
      f[i][j] = f[p][j - 1];
    }
    for (i = 1; i <= n; ++i) f[i][j] += f[i - 1][j], f[i][j] %= md;
  }
  return f[n][k];
}
int main() {
  scanf("%d%d", &n, &k);
  int i;
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  int p = a[n] / (k - 1);
  dp[p] = sol(p);
  ans = (long long)p * (dp[p]) % md;
  for (i = p - 1; i >= 1; --i) {
    dp[i] = sol(i);
    ans += (long long)i * (dp[i] - dp[i + 1] + md) % md;
    ans %= md;
  }
  cout << ans;
}
