#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long k, n, b[1005], ans;
long long dp[1005][1005];
long long res[100005];
long long sum[1005][1005];
int main() {
  scanf("%d%d", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%d", &b[i]);
  sort(b + 1, b + n + 1);
  b[0] = -1000000000;
  for (long long v = 1; v * (k - 1) <= b[n]; v++) {
    dp[0][0] = 1, sum[0][0] = 1;
    long long now = 0;
    for (long long i = 1; i <= n; i++) {
      while (b[now] <= b[i] - v) now++;
      for (long long j = 0; j <= k; j++) {
        if (j) dp[i][j] = sum[now - 1][j - 1];
        sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
      }
      res[v] = (res[v] + dp[i][k]) % mod;
    }
    ans = (ans + res[v]) % mod;
  }
  printf("%lld\n", ans % mod);
  return 0;
}

### Prompt

Please formulate 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;
const long long mod = 998244353;
long long k, n, b[1005], ans;
long long dp[1005][1005];
long long res[100005];
long long sum[1005][1005];
int main() {
  scanf("%d%d", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%d", &b[i]);
  sort(b + 1, b + n + 1);
  b[0] = -1000000000;
  for (long long v = 1; v * (k - 1) <= b[n]; v++) {
    dp[0][0] = 1, sum[0][0] = 1;
    long long now = 0;
    for (long long i = 1; i <= n; i++) {
      while (b[now] <= b[i] - v) now++;
      for (long long j = 0; j <= k; j++) {
        if (j) dp[i][j] = sum[now - 1][j - 1];
        sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
      }
      res[v] = (res[v] + dp[i][k]) % mod;
    }
    ans = (ans + res[v]) % mod;
  }
  printf("%lld\n", ans % mod);
  return 0;
}

```