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;
vector<pair<int, int>> v[100004];
int k, n, i, max_diff, diff, mod, j, h, f, l, x, y, a[1004], dp[3][1004][1004],
    s[3][1004][1004];
long long sol;
int main() {
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  max_diff = (a[n] - a[1]) / (k - 1);
  for (i = 1; i <= n; i++)
    for (j = (i + 1); j <= n && (a[j] - a[i]) <= max_diff; j++)
      v[a[j] - a[i]].push_back({i, j});
  sol = 0LL;
  mod = 998244353;
  for (diff = 1; diff <= max_diff; diff++) {
    if ((int)v[diff].size() == 0) continue;
    for (j = 1; j <= n; j++)
      for (h = 1; h <= k; h++) {
        dp[1][j][h] = 0;
        dp[2][j][h] = 0;
        s[1][j][h] = 0;
        s[2][j][h] = 0;
      }
    for (j = 1; j <= n; j++) {
      dp[1][j][1] = 1;
      s[1][j][1] = j;
      dp[2][j][1] = 1;
      s[2][j][1] = n - j + 1;
    }
    for (h = 1; j <= k; h++) {
      dp[1][0][h] = 0;
      s[1][0][h] = 0;
      dp[2][n + 1][h] = 0;
      s[2][n + 1][h] = 0;
    }
    for (h = 2; h <= k; h++) {
      l = 1;
      for (j = 1; j <= n; j++) {
        while ((a[j] - a[l]) > diff) l++;
        if (l > 1)
          dp[1][j][h] = s[1][l - 1][h - 1];
        else
          dp[1][j][h] = 0;
        s[1][j][h] = (s[1][j - 1][h] + dp[1][j][h]) % mod;
      }
    }
    for (h = 2; h <= k; h++) {
      l = n;
      for (j = n; j >= 1; j--) {
        while ((a[l] - a[j]) >= diff) l--;
        if (l < n)
          dp[2][j][h] = s[2][l + 1][h - 1];
        else
          dp[2][j][h] = 0;
        s[2][j][h] = (s[2][j + 1][h] + dp[2][j][h]) % mod;
      }
    }
    for (j = 0; j < (int)v[diff].size(); j++) {
      x = v[diff][j].first;
      y = v[diff][j].second;
      l = min(k - 1, ((a[x] - a[1]) / diff) + 1);
      f = max(1, k - ((a[n] - a[y]) / diff) - 1);
      for (h = f; h <= l; h++) {
        sol = (0LL) + sol +
              (1LL) * dp[1][x][h] * dp[2][y][k - h] % mod * diff % mod;
        sol = (1LL) * sol % mod;
      }
    }
  }
  printf("%I64d\n", sol);
  return 0;
}
