#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
const int mod = 998244353;
const int N = 1005;
int dp[N][N], pre[N][N];
int a[N];
int n, k;
int solve(int x) {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j] = 0;
    }
  }
  int pt = 0;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    dp[i][1] = 1;
    while (pt + 1 < i && a[pt + 1] + x <= a[i]) pt++;
    if (a[i] - a[pt] >= x) {
      for (int j = 2; j <= k; j++) {
        dp[i][j] = pre[pt][j - 1];
      }
    }
    for (int j = 1; j <= k; j++) {
      pre[i][j] = (pre[i - 1][j] + dp[i][j]) % mod;
    }
    ans = (ans + dp[i][k]) % mod;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int x = 1; x <= a[n] / (k - 1); x++) {
    ans = (ans + solve(x)) % mod;
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
const int mod = 998244353;
const int N = 1005;
int dp[N][N], pre[N][N];
int a[N];
int n, k;
int solve(int x) {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j] = 0;
    }
  }
  int pt = 0;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    dp[i][1] = 1;
    while (pt + 1 < i && a[pt + 1] + x <= a[i]) pt++;
    if (a[i] - a[pt] >= x) {
      for (int j = 2; j <= k; j++) {
        dp[i][j] = pre[pt][j - 1];
      }
    }
    for (int j = 1; j <= k; j++) {
      pre[i][j] = (pre[i - 1][j] + dp[i][j]) % mod;
    }
    ans = (ans + dp[i][k]) % mod;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int x = 1; x <= a[n] / (k - 1); x++) {
    ans = (ans + solve(x)) % mod;
  }
  cout << ans;
  return 0;
}

```