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 = 1111, mod = 998244353;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 0;
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n, k, a[maxn], f[maxn][maxn], s[maxn][maxn], ans;
int main() {
  n = read();
  k = read();
  for (int i = (1); i <= (n); i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int x = (1); x <= ((a[n] - a[1]) / (k - 1)); x++) {
    for (int i = (0); i <= (n); i++)
      for (int j = (0); j <= (k); j++) f[i][j] = s[i][j] = 0;
    for (int i = (1); i <= (n); i++) f[i][1] = 1, s[i][1] = i;
    for (int j = (2); j <= (k); j++) {
      int cur = 0;
      for (int i = (1); i <= (n); i++) {
        while (cur < i && a[i] - a[cur] >= x) cur++;
        if (cur && a[i] - a[cur] < x) cur--;
        f[i][j] = s[cur][j - 1];
        s[i][j] = (s[i - 1][j] + f[i][j]) % mod;
      }
    }
    ans = (ans + s[n][k]) % mod;
  }
  printf("%d\n", ans);
}
