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 maxn = 1e3 + 10;
const int maxA = 1e5 + 10;
const int mod = 998244353;
int a[maxn], dp[maxn], par[maxn], pnt[maxn], pd[maxA];
int n, k;
int mp[maxA];
int cal(int x) {
  if (mp[x] != -1) return mp[x];
  int ptr = 0;
  for (int i = 1; i <= n; i++) {
    while (ptr < i - 1 and a[i] - a[ptr + 1] >= x) ptr++;
    pnt[i] = ptr;
  }
  for (int i = 0; i <= n; i++) dp[i] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = n; j >= 1; j--) dp[j] = dp[pnt[j]];
    dp[0] = 0;
    for (int j = 1; j <= n; j++) {
      dp[j] += dp[j - 1];
      if (dp[j] >= mod) dp[j] -= mod;
    }
  }
  return mp[x] = dp[n];
}
int main() {
  memset(mp, -1, sizeof mp);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[0] = -a[n];
  sort(a + 1, a + n + 1);
  int now = 0;
  vector<pair<int, int> > v;
  while (now <= a[n]) {
    int t = cal(now);
    int lo = now, hi = a[n] + 1;
    while (hi - lo > 1) {
      int mid = (lo + hi) >> 1;
      if (cal(mid) == t)
        lo = mid;
      else
        hi = mid;
    }
    v.push_back({lo, t});
    now = hi;
  }
  int m = v.size();
  int answer = 0;
  for (int i = 0; i < m; i++) {
    int t = v[i].second;
    if (i < m - 1) t = (t - v[i + 1].second + mod) % mod;
    answer = (answer + 1ll * t * v[i].first % mod) % mod;
  }
  printf("%d\n", answer);
}
