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>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
using namespace ::std;
const int maxn = 1100;
const int inf = 1e9 + 800;
const int mod = 998244353;
void jam(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int jamO(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int a[maxn];
int dp[2][maxn];
int par[2][maxn];
int f[maxn];
int findHal(int n, int k, int m) {
  int p = -1;
  for (int i = 0; i < n; i++) {
    while (a[p + 1] <= a[i] - m) p++;
    f[i] = p;
  }
  for (int i = 0; i < n; i++) {
    dp[1][i] = 1;
  }
  par[1][0] = dp[1][0];
  for (int i = 1; i < n; i++) {
    par[1][i] = jamO(par[1][i - 1], dp[1][i]);
  }
  for (int j = 2; j <= k; j++) {
    bool noww = (j & 1);
    bool lass = !noww;
    for (int i = 0; i < n; i++) {
      if (f[i] != -1)
        dp[noww][i] = par[lass][f[i]];
      else
        dp[noww][i] = 0;
    }
    par[noww][0] = dp[noww][0];
    for (int i = 1; i < n; i++) {
      par[noww][i] = jamO(par[noww][i - 1], dp[noww][i]);
    }
  }
  return par[k & 1][n - 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int ans = 0;
  int mac = (1e5) / (k - 1) + 2;
  for (int i = 1; i <= mac; i++) {
    jam(ans, findHal(n, k, i));
  }
  cout << ans;
}
