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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long int rand(long long int x, long long int y) {
  ++y;
  return (rng() % (y - x)) + x;
}
string to_string(char c) {
  string second(1, c);
  return second;
}
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void degug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void degug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  degug_out(T...);
}
inline long long int gcd(long long int a, long long int b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int n, k, A[(1006)], ans, psum[(1006)];
vector<long long int> dp[(1006)];
const long long int MOD = 998244353;
long long int solve(long long int x) {
  long long int co = 1;
  long long int ans = 0;
  for (auto &i : dp)
    for (auto &j : i) j = 0;
  memset((psum), 0, sizeof((psum)));
  dp[0][0] = 1;
  psum[0] += dp[0][0];
  for (long long int i = (1); i < (long long int)(n + 1); ++i) {
    while (co < i && A[co] + x <= A[i]) {
      for (long long int i = (0); i < (long long int)(k + 1); ++i)
        psum[i] += dp[co][i], psum[i] %= MOD;
      ++co;
    }
    for (long long int j = (1); j < (long long int)(k + 1); ++j) {
      dp[i][j] += psum[j - 1];
      dp[i][j] %= MOD;
    }
    ans += dp[i][k];
    ans += MOD;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (long long int i = (1); i < (long long int)(n + 1); ++i) cin >> A[i];
  sort(A + 1, A + n + 1);
  long long int mx = *max_element(A + 1, A + n + 1);
  for (long long int i = (0); i < (long long int)((1006)); ++i)
    dp[i].resize(k + 2);
  for (long long int x = (1); x < (long long int)(mx + 1); ++x)
    if (x * (k - 1) <= mx) {
      ans += solve(x);
      ans %= MOD;
    }
  cout << ans << '\n';
}
