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
var n = parseInt(readline())
var p = []
var q = []
for (i = 0; i < n; i++) {
  var line = readline().split(' ')
  p[i] = {x:parseInt(line[0]), y:parseInt(line[1])}
  q[i] = p[i].x
  q[i+n] = p[i].y
}
p.sort(function(a,b) { return (b.x-b.y) - (a.x-a.y)})
var ans = 0
for (aa in q) {
  var a=q[aa]
  var now = 0
  var ok = true
  for (bb in p) {
    b=p[bb]
    if (b.y > a && b.x > a) {
      ok = false
      break
    }
    if (b.y > a ) {
      now++
    }
  }
  if (!ok) {
    continue
  }
  if (now>n/2) {
    continue
  }
  var nowans = 0
  var ms = []
  for (bb in p) {
    b = p[bb]
    if (now+1 <= n/2 && b.x <= a && b.y < b.x) {
      now++;
      nowans+=b.y
      ms.push(b)
    } else if (b.y>a){
      nowans+=b.y
    } else {
      nowans+=b.x
    }
  }
  if (ans <=0 || (nowans*a < ans)) {
    ans = nowans*a
  }
}
print(ans)
