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>
const int maxn = 1e3 + 23;
const int mod = 998244353;
using namespace std;
void read(int &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
int T, n, m, q, k;
long long f[maxn][maxn];
int a[maxn];
int main() {
  read(n);
  read(k);
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + n + 1);
  int inf = a[n] / (k - 1);
  long long ans = 0;
  for (int i = 1; i <= inf; i++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= k; j++) f[i][j] = 0;
    for (int i = 1; i <= n; i++) f[i][0] = 1;
    f[0][0] = 1;
    f[1][0] = f[1][1] = 1;
    int j = 0, p = 1;
    for (j = 2; j <= n; j++) {
      for (; p <= j; p++) {
        if (a[j] - a[p] < i) {
          p--;
          break;
        }
      }
      p = max(p, 0);
      for (int l = 1; l <= min(k, j); l++) {
        f[j][l] = f[j - 1][l] + f[p][l - 1];
        f[j][l] %= mod;
      }
    }
    ans += f[n][k];
    ans %= mod;
  }
  printf("%lld", ans);
}
