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 = 1005;
const int MAXV = 100005;
const long long MOD = 998244353;
int n, m;
int a[MAXN];
long long ans;
long long cnt[MAXV];
long long f[MAXN][MAXN];
void calc(int v) {
  for (int i = 1; i <= n; i++) f[1][i] = i;
  for (int i = 2; i <= m; i++) {
    for (int j = 1, k = 0; j <= n; j++) {
      while (k + 1 < j && a[j] - a[k + 1] >= v) k++;
      f[i][j] = f[i - 1][k];
    }
    for (int j = 1; j <= n; j++) (f[i][j] += f[i][j - 1]) %= MOD;
  }
  cnt[v] = f[m][n];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 0; i * (m - 1) <= 100000; i++) calc(i);
  for (int i = 0; i * (m - 1) <= 100000; i++)
    (ans += i * (cnt[i] - cnt[i + 1])) %= MOD;
  printf("%lld\n", (ans + MOD) % MOD);
  return 0;
}
