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;
inline int read() {
  int x = 0, k = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * k;
}
const int mod = 998244353;
int n, k, cnt = 0;
long long ans = 0;
int a[1005], dp[1005][1005];
inline long long get(int val) {
  dp[0][0] = 1;
  int now = 0;
  for (register int i = 1; i <= n; ++i) {
    while (a[i] - a[now + 1] >= val) ++now;
    dp[i][0] = 1;
    for (register int j = 1; j <= k; ++j)
      dp[i][j] = (dp[i - 1][j] + dp[now][j - 1]) % mod;
  }
  return dp[n][k];
}
int main() {
  n = read(), k = read();
  for (register int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  a[0] = -1;
  for (register int i = 1; i * (k - 1) <= a[n]; ++i) ans = (ans + get(i)) % mod;
  return printf("%lld", ans) & 0;
}
