Many years have passed, and n friends met at a party again. Technologies have leaped forward since the last meeting, cameras with timer appeared and now it is not obligatory for one of the friends to stand with a camera, and, thus, being absent on the photo.

Simply speaking, the process of photographing can be described as follows. Each friend occupies a rectangle of pixels on the photo: the i-th of them in a standing state occupies a wi pixels wide and a hi pixels high rectangle. But also, each person can lie down for the photo, and then he will occupy a hi pixels wide and a wi pixels high rectangle.

The total photo will have size W × H, where W is the total width of all the people rectangles, and H is the maximum of the heights. The friends want to determine what minimum area the group photo can they obtain if no more than n / 2 of them can lie on the ground (it would be strange if more than n / 2 gentlemen lie on the ground together, isn't it?..)

Help them to achieve this goal.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of friends.

The next n lines have two integers wi, hi (1 ≤ wi, hi ≤ 1000) each, representing the size of the rectangle, corresponding to the i-th friend.

Output

Print a single integer equal to the minimum possible area of the photo containing all friends if no more than n / 2 of them can lie on the ground.

Examples

Input

3
10 1
20 2
30 3


Output

180


Input

3
3 1
2 2
4 3


Output

21


Input

1
5 10


Output

50
#include <bits/stdc++.h>
using namespace std;
int X = 1000000005, H[1005], W[1005], N;
int g(int h, int k, int w) {
  priority_queue<int> q;
  for (int i = 1; i <= N; i++) {
    if (H[i] <= h && W[i] <= h) {
      q.push(W[i] - H[i]);
      w += W[i];
    } else if (W[i] <= h && k) {
      w += H[i];
      k--;
    } else if (H[i] <= h)
      w += W[i];
    else
      return 1000000005;
  }
  while (!q.empty() && k && q.top() > 0) {
    w -= q.top();
    q.pop();
    k--;
  }
  return w * h;
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> W[i] >> H[i];
  for (int h = 1; h <= 1000; h++) X = min(X, g(h, N / 2, 0));
  cout << X << endl;
  return 0;
}
