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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
const int MN = 1005, MA = 100005, inf = 1000000005, mod = 998244353;
const long long INF = 1000000000000000005LL;
int B[MN];
long long dp[MN][MN];
int n, k;
long long licz_dp(int b) {
  for (int i = 1; i <= n; ++i) dp[1][i] = 1LL;
  for (int l = 2; l <= k; ++l) {
    long long suma = 0LL;
    int wsk = 1;
    for (int i = 1; i <= n; ++i) {
      while (wsk < i && B[i] - B[wsk] >= b) suma += dp[l - 1][wsk++];
      dp[l][i] = suma % mod;
    }
  }
  long long ret = 0LL;
  for (int i = 1; i <= n; ++i) ret += dp[k][i];
  return ret % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &B[i]);
  sort(B + 1, B + 1 + n);
  int bound = 1e5 / (k - 1) + 3;
  long long ans = 0LL;
  for (int i = 1; i <= bound; ++i) ans += licz_dp(i);
  printf("%lld", ans % mod);
}
