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());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 998244353ll;
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    ex >>= 1ll;
    x = (x * x) % mod;
  }
  return ans;
}
const int maxn = 1e6 + 7;
const long long inf = 1e9 + 7;
long long a[maxn];
long long sol[maxn];
int nx[maxn];
long long suf[maxn];
long long dp[maxn];
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n, k;
  cin >> n >> k;
  for (int j = 1; j <= n; j++) cin >> a[j];
  sort(a + 1, a + n + 1);
  ;
  fill(sol, sol + maxn, 0);
  for (long long x = 1;; x++) {
    ;
    int p = 1;
    for (int j = 1; j <= n; j++) {
      while (p <= n && a[p] < a[j] + x) p++;
      nx[j] = p;
      ;
    }
    for (int j = 1; j <= n; j++) dp[j] = 1;
    for (int t = 2; t <= k; t++) {
      suf[n + 1] = 0ll;
      for (int j = n; j >= 1; j--) {
        suf[j] = (suf[j + 1] + dp[j]) % mod;
      }
      for (int j = 1; j <= n; j++) {
        dp[j] = suf[nx[j]];
        if (k == t) sol[x] += dp[j];
      }
    }
    sol[x] %= mod;
    if (sol[x] == 0) break;
    ;
  }
  long long ans = 0ll;
  for (long long x = 1;; x++) {
    if (sol[x] == 0) break;
    long long ca = (x * (sol[x] - sol[x + 1] + mod)) % mod;
    ans += ca;
  }
  ans %= mod;
  cout << ans << "\n";
}
