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 long long maxn = 1010;
const long long mod = 998244353;
inline long long sum(long long a, long long b) { return (a + b) % mod; }
long long n, k;
long long a[maxn];
long long dp[maxn][maxn], pref[maxn][maxn];
long long le[maxn];
long long solve(long long val) {
  for (long long i = 0; i <= k; i++)
    for (long long j = 0; j <= n; j++) dp[i][j] = 0, pref[i][j] = 0;
  le[0] = 0;
  for (long long i = 1; i <= n; i++) {
    le[i] = le[i - 1];
    while (a[i] - a[le[i] + 1] >= val) le[i]++;
  }
  dp[0][0] = 1;
  for (long long i = 0; i + 1 <= k; i++) {
    pref[i][0] = dp[i][0];
    for (long long j = 1; j <= n; j++) {
      pref[i][j] = sum(pref[i][j - 1], dp[i][j]);
    }
    for (long long j = 1; j <= n; j++) {
      dp[i + 1][j] = pref[i][le[j]];
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans = sum(ans, dp[k][i]);
  return ans;
}
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (long long i = 1; i * (k - 1) <= 100000 + 10; i++)
    ans = sum(ans, solve(i));
  cout << ans;
}
