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 = 1010;
const int INF = 0x3f3f3f3f;
const long long mod = 998244353;
int a[maxn], n, m;
long long dp[maxn][maxn];
long long Cal(int x) {
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= m; ++j) dp[i][j] = 0;
  dp[0][0] = 1;
  for (int j = 1; j <= m; ++j) {
    long long pre = 0;
    for (int i = 1, k = 0; i <= n; ++i) {
      while (k < i && a[i] - a[k] >= x) {
        pre += dp[k][j - 1];
        k++;
      }
      dp[i][j] += pre;
      dp[i][j] %= mod;
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; ++i) res += dp[i][m];
  return res % mod;
}
int main() {
  scanf("%d %d", &n, &m);
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    mx = max(mx, a[i]);
  }
  a[0] = -INF;
  sort(a, a + 1 + n);
  long long ans = 0;
  mx /= (m - 1);
  for (int i = 1; i <= mx; ++i) ans += Cal(i);
  printf("%lld\n", ans % mod);
}
