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 = 1020;
const int maxm = 1000020;
const int MOd = 998244353;
int a, b, ar[maxn], dn[maxn][maxn];
int nex[maxn], val[maxn * 100];
int add(int a, int b) {
  a += b;
  a %= MOd;
  if (a < 0) return a + MOd;
  return a;
}
int mul(int a, int b) { return (long long)a * b % MOd; }
int us(int x, int y) {
  int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
int inv(int x) { return us(x, MOd - 2); }
int f(int x) {
  for (int i = 1, j = 1; i <= a; i++) {
    while (j <= a && ar[j] < ar[i] + x) j++;
    nex[i] = j;
  }
  dn[a + 1][0] = 1;
  for (int i = a; i >= 1; i--)
    for (int j = b; j >= 0; j--) {
      dn[i][j] = dn[i + 1][j];
      if (j) dn[i][j] = add(dn[i][j], dn[nex[i]][j - 1]);
    }
  return dn[1][b];
}
int main() {
  scanf("%d %d", &a, &b);
  for (int i = 1; i <= a; i++) {
    scanf("%d", &ar[i]);
  }
  sort(ar + 1, ar + 1 + a);
  int ans = 0;
  for (int i = 1; i <= 1e5 + 1; i++) {
    val[i] = f(i);
    if (i > 1) ans = add(ans, mul(i - 1, add(val[i - 1], -val[i])));
    if (val[i] == 0) break;
  }
  cout << ans << endl;
  return 0;
}
