#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int sum(int a, int b) {
  int s = (a + b);
  if (s >= mod) s -= mod;
  return s;
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
const int maxN = 1005;
int a[maxN];
int n, k;
int dp[maxN][maxN];
int pref[maxN];
int le[maxN];
int solve(int val) {
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = 0;
    }
  }
  le[0] = 0;
  for (int i = 1; i <= n; i++) {
    le[i] = le[i - 1];
    while (a[i] - a[le[i] + 1] >= val) le[i]++;
  }
  dp[0][0] = 1;
  for (int i = 0; i + 1 <= k; i++) {
    pref[0] = dp[i][0];
    for (int j = 1; j <= n; j++) {
      pref[j] = sum(pref[j - 1], dp[i][j]);
    }
    for (int j = 1; j <= n; j++) {
      dp[i + 1][j] = pref[le[j]];
    }
  }
  int f = 0;
  for (int i = 1; i <= n; i++) f = sum(f, dp[k][i]);
  return f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int a = 1; a * (k - 1) <= 100000 + 10; a++) {
    ans = sum(ans, solve(a));
  }
  cout << 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 int mod = 998244353;
int sum(int a, int b) {
  int s = (a + b);
  if (s >= mod) s -= mod;
  return s;
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
const int maxN = 1005;
int a[maxN];
int n, k;
int dp[maxN][maxN];
int pref[maxN];
int le[maxN];
int solve(int val) {
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = 0;
    }
  }
  le[0] = 0;
  for (int i = 1; i <= n; i++) {
    le[i] = le[i - 1];
    while (a[i] - a[le[i] + 1] >= val) le[i]++;
  }
  dp[0][0] = 1;
  for (int i = 0; i + 1 <= k; i++) {
    pref[0] = dp[i][0];
    for (int j = 1; j <= n; j++) {
      pref[j] = sum(pref[j - 1], dp[i][j]);
    }
    for (int j = 1; j <= n; j++) {
      dp[i + 1][j] = pref[le[j]];
    }
  }
  int f = 0;
  for (int i = 1; i <= n; i++) f = sum(f, dp[k][i]);
  return f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int a = 1; a * (k - 1) <= 100000 + 10; a++) {
    ans = sum(ans, solve(a));
  }
  cout << ans;
  return 0;
}

```