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 P = 998244353;
int n, k;
int dp[2222][1111], a[2222];
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
int main() {
  read(n);
  read(k);
  for (int i = 1; i <= n; ++i) read(a[i]);
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = (int)1e5 / (k - 1); i >= 1; --i) {
    int pt = 1;
    for (int j = 1; j <= n; ++j) {
      while (pt < j && a[j] - a[pt] >= i) ++pt;
      dp[j][2] = (dp[j - 1][2] + pt - 1) % P;
      for (int K = 3; K <= k; ++K)
        dp[j][K] = (1ll * dp[j - 1][K] + dp[pt - 1][K - 1]) % P;
    }
    ans = (ans + dp[n][k]) % P;
  }
  printf("%d\n", ans);
  return 0;
}
