#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const int mod = 998244353;
int dp[N][N];
int pre[N];
int a[N];
int n, k, ans;
void prep(int i) {
  for (int j = 1; j <= n; j++) {
    while (pre[j] > 0 && a[j] - a[pre[j]] < i) {
      pre[j]--;
    }
  }
}
void update() {
  for (int m = 1; m <= k; m++) {
    for (int j = 1; j <= n; j++) {
      dp[m][j] = (dp[m][j - 1] + dp[m - 1][pre[j]]) % mod;
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 0; i <= n; i++) {
    dp[0][i] = 1;
  }
  for (int i = 1; i * (k - 1) + a[1] <= a[n]; i++) {
    prep(i);
    update();
    ans = (ans + dp[k][n]) % mod;
  }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 1005;
const int mod = 998244353;
int dp[N][N];
int pre[N];
int a[N];
int n, k, ans;
void prep(int i) {
  for (int j = 1; j <= n; j++) {
    while (pre[j] > 0 && a[j] - a[pre[j]] < i) {
      pre[j]--;
    }
  }
}
void update() {
  for (int m = 1; m <= k; m++) {
    for (int j = 1; j <= n; j++) {
      dp[m][j] = (dp[m][j - 1] + dp[m - 1][pre[j]]) % mod;
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 0; i <= n; i++) {
    dp[0][i] = 1;
  }
  for (int i = 1; i * (k - 1) + a[1] <= a[n]; i++) {
    prep(i);
    update();
    ans = (ans + dp[k][n]) % mod;
  }
  cout << ans << endl;
}

```