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 = 1007;
const int MX = 998244353;
const long long int INF = 1LL * MX * MX;
int n, k, ans;
int in[N];
int dp[N][N];
void add(int &a, int b) {
  a += b;
  if (a >= MX) a -= MX;
}
int solve2(int t) {
  in[0] = -MX;
  int last = 0, ret = 0;
  for (int i = 1; i <= n; ++i)
    if (in[i] - in[last] >= t) ++ret, last = i;
  return ret >= k;
}
int solve(int t) {
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= k; ++j) dp[i][j] = 0;
  int last = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    while (in[last + 1] + t <= in[i]) ++last;
    dp[i][0] = dp[i - 1][0];
    for (int j = 1; j <= k; ++j) {
      dp[i][j] = dp[i - 1][j];
      add(dp[i][j], dp[last][j - 1]);
    }
  }
  return dp[n][k];
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &in[i]);
  sort(in + 1, in + n + 1);
  set<int> sens;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) sens.insert(in[i] - in[j]);
  int s = 0, e = 100 * 1000;
  while (s < e) {
    int mid = (s + e + 1) >> 1;
    if (solve2(mid) == 0)
      e = mid - 1;
    else
      s = mid;
  }
  int pref = 0;
  for (auto v : sens) {
    if (-v > s) continue;
    if (v == 0) break;
    int cur = solve(-v);
    ans = (ans + 1LL * (cur - pref + MX) * (-v)) % MX;
    pref = cur;
  }
  printf("%d\n", ans);
  return 0;
}
