#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int maxn = 1e3 + 5;
int n, k;
long long dp[maxn][maxn];
int m = 1e5;
int a[maxn];
long long suf[maxn][maxn];
int main() {
  scanf("%d%d", &n, &k);
  int mn = m / (k - 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -1e9;
  long long ans = 0;
  for (int w = 1; w <= mn; w++) {
    int lb = 0;
    suf[0][0] = 1;
    for (int j = 1; j <= k; j++) suf[0][j] = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[lb] >= w) lb++;
      lb--;
      suf[i][0] = 1;
      for (int j = 1; j <= k; j++) {
        suf[i][j] = suf[lb][j - 1] + suf[i - 1][j], suf[i][j] %= mod;
      }
    }
    ans += suf[n][k];
    ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp 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 long long mod = 998244353;
const int maxn = 1e3 + 5;
int n, k;
long long dp[maxn][maxn];
int m = 1e5;
int a[maxn];
long long suf[maxn][maxn];
int main() {
  scanf("%d%d", &n, &k);
  int mn = m / (k - 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -1e9;
  long long ans = 0;
  for (int w = 1; w <= mn; w++) {
    int lb = 0;
    suf[0][0] = 1;
    for (int j = 1; j <= k; j++) suf[0][j] = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[lb] >= w) lb++;
      lb--;
      suf[i][0] = 1;
      for (int j = 1; j <= k; j++) {
        suf[i][j] = suf[lb][j - 1] + suf[i - 1][j], suf[i][j] %= mod;
      }
    }
    ans += suf[n][k];
    ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```