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.
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long MNOGO = 2e9;
const long long MD = 998244353;
const double EPS = 1e-7;
struct pryama {
  double a, b, c;
};
class segmentTree {
 public:
  segmentTree* left = nullptr;
  segmentTree* right = nullptr;
  long long tl = 0, tr = 1 << 22;
  long long val = 0;
  void createSons() {
    if (left == nullptr) {
      left = new segmentTree();
      left->tl = tl;
      left->tr = (tl + tr) / 2;
    }
    if (right == nullptr) {
      right = new segmentTree();
      right->tr = tr;
      right->tl = (tl + tr) / 2 + 1;
    }
  }
  pair<long long, long long> merge(pair<long long, long long> l,
                                   pair<long long, long long> r) {
    if (l.first == r.first) {
      return {l.first, r.second};
    }
    if (l.first > r.first) {
      return l;
    } else {
      return r;
    }
  };
  void update(long long pos, long long q) {
    if (tl == tr) {
      val = q;
      return;
    }
    createSons();
    long long md = (tl + tr) / 2;
    if (pos <= md) {
      left->update(pos, q);
    } else {
      right->update(pos, q);
    }
    val = min(left->val, right->val);
  }
  long long get(long long l, long long r) {
    if (l > r) {
      return 0;
    }
    if (l > tr || r < tl) {
      return 0;
    }
    if (tl >= l && tr <= r) {
      return val;
    }
    createSons();
    return min(left->get(l, r), right->get(l, r));
  }
  void build(long long l, long long r, long long* a) {
    if (tl > tr) {
      return;
    }
    if (l > tr || r < tl) {
      return;
    }
    if (tl == tr) {
      val = a[tl];
      return;
    }
    createSons();
    left->build(l, r, a);
    right->build(l, r, a);
    val = min(left->get(l, r), right->get(l, r));
  }
};
long long n, k;
long long a[1000006];
long long dp[1003][1003];
long long sum[1003][1003];
long long p[100003];
long long solve(long long x) {
  a[0] = -INF;
  long long uk1 = 1;
  for (long long i = 1; i <= n; i++) {
    while (a[uk1] + x <= a[i]) {
      uk1++;
    }
    for (long long j = 1; j <= k; j++) {
      long long pos = uk1;
      if (j == 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] = sum[pos - 1][j - 1];
      }
      sum[i][j] = (dp[i][j] + sum[i - 1][j]) % MD;
    }
  }
  return sum[n][k];
}
long long ans = 0;
void get_all(long long fr) {
  for (long long i = fr; i >= 1; i--) {
    p[i] = solve(i);
    if (p[i] != 0) {
      ans = (ans + ((p[i] + MD - p[i + 1]) % MD * i) % MD) % MD;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  get_all(100000 / (k - 1) + 1);
  cout << ans << '\n';
}
