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;
template <class T>
using v2d = vector<vector<T> >;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 1e3 + 10;
const int maxK = 1e3 + 10;
const long long mod = 998244353;
int n, k, a[maxN], lowbound[maxN], upbound[maxN];
long long dp[maxN][maxK], res = 0;
inline void normalize(long long &a) { a = (a % mod + mod) % mod; }
long long Try(int dist) {
  long long r = 0;
  for (int i = 1; i <= (int)(n); ++i) {
    lowbound[i] = lowbound[i - 1];
    while (a[i] - a[lowbound[i] + 1] >= dist) {
      ++lowbound[i];
    }
    upbound[i] = upbound[i - 1];
    while (a[i] - a[upbound[i] + 1] > dist) {
      ++upbound[i];
    }
  }
  for (int i = 1; i <= (int)(n); ++i) {
    dp[i][1] = dp[i - 1][1] + 1;
    for (int j = (int)(2); j <= (int)(k); ++j) {
      dp[i][j] = dp[lowbound[i]][j - 1] + dp[i - 1][j];
      if (dp[i][j] >= mod) {
        dp[i][j] -= mod;
      }
    }
  }
  r = dp[n][k];
  for (int i = 1; i <= (int)(n); ++i) {
    dp[i][1] = dp[i - 1][1] + 1;
    for (int j = (int)(2); j <= (int)(k); ++j) {
      dp[i][j] = dp[upbound[i]][j - 1] + dp[i - 1][j];
      if (dp[i][j] >= mod) {
        dp[i][j] -= mod;
      }
    }
  }
  r -= dp[n][k];
  if (r < 0) {
    r += mod;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= (int)(n); ++i) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= (int)(a[n] / (k - 1)); ++i) {
    res += i * Try(i) % mod;
    if (res >= mod) {
      res -= mod;
    }
  }
  cout << res;
  return 0;
}
