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;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modularExponentiation(long long base, long long power,
                                long long mod) {
  long long ans = 1;
  while (power > 0) {
    if (power & 1) ans = (ans * base) % mod;
    power = power >> 1;
    base = (base * base) % mod;
  }
  return ans;
}
const int N = 1010;
int n, k, arr[N], le[N];
long long pref[N][N], dp[N][N];
long long calculate(int a) {
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= k; ++j) dp[i][j] = 0;
  le[0] = 0;
  for (int i = 1; i <= n; ++i) {
    le[i] = le[i - 1];
    while (arr[i] - arr[le[i] + 1] >= a) le[i]++;
  }
  dp[0][0] = pref[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    pref[i][0] = 1;
    for (int j = 1; j <= min(i, k); ++j) {
      dp[i][j] = pref[le[i]][j - 1];
    }
    for (int j = 1; j <= min(i, k); ++j) {
      pref[i][j] = (pref[i - 1][j] + dp[i][j]) % 998244353;
    }
  }
  long long ret = 0;
  for (int i = 1; i <= n; ++i) ret = (ret + dp[i][k]) % 998244353;
  return ret;
}
void solve(int test) {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  sort(arr + 1, arr + n + 1);
  long long ans = 0;
  for (int a = 1; a * (k - 1) <= 100010; ++a) {
    ans = (ans + calculate(a)) % 998244353;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; ++i) {
    solve(i);
  }
  return 0;
}
