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 INF = 0x3c3c3c3c;
const long long LINF = 1ll * INF * INF * 2;
const int MOD = 998244353;
const int N = 1010;
int n, m;
int x[N];
long long cnt[N][N];
long long ans[100100];
long long f(int k) {
  for (int i = 0; i < n; i++) cnt[i][1] = 1;
  for (int i = n - 2; i >= 0; i--)
    cnt[i][1] = (cnt[i][1] + cnt[i + 1][1]) % MOD;
  for (int t = 2; t <= m; t++) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      while (cur <= i || (cur < n && x[cur] < x[i] + k)) cur++;
      cnt[i][t] = cnt[cur][t - 1];
    }
    for (int i = n - 2; i >= 0; i--)
      cnt[i][t] = (cnt[i][t] + cnt[i + 1][t]) % MOD;
  }
  return cnt[0][m];
}
int main(void) {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", x + i);
  sort(x, x + n);
  int M = 100000 / (m - 1);
  for (int k = 0; k <= M; k++) ans[k] = f(k);
  long long sum = 0;
  for (int k = 0; k <= M; k++) sum = (sum + k * (ans[k] - ans[k + 1])) % MOD;
  printf("%lld", (sum % MOD + MOD) % MOD);
}
