PROBLEM LINK ::http://codeforces.com/contest/205/problem/D

D. Little Elephant and Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Little Elephant loves to play with color cards.

He has n cards, each has exactly two colors (the color of the front side and the color of the back side). Initially, all the cards lay on the table with the front side up. In one move the Little Elephant can turn any card to the other side. The Little Elephant thinks that a set of cards on the table is funny if at least half of the cards have the same color (for each card the color of the upper side is considered).

Help the Little Elephant to find the minimum number of moves needed to make the set of n cards funny.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of the cards. The following n lines contain the description of all cards, one card per line. The cards are described by a pair of positive integers not exceeding 109 — colors of both sides. The first number in a line is the color of the front of the card, the second one — of the back. The color of the front of the card may coincide with the color of the back of the card.

The numbers in the lines are separated by single spaces.

Output

On a single line print a single integer — the sought minimum number of moves. If it is impossible to make the set funny, print -1.

Sample test(s)

input

3
4 7
4 7
7 4

output

0

input

5
4 7
7 4
2 11
9 7
1 1

output

2

Note

In the first sample there initially are three cards lying with colors 4, 4, 7. Since two of the three cards are of the same color 4, you do not need to change anything, so the answer is 0.

In the second sample, you can turn the first and the fourth cards. After that three of the five cards will be of color 7.

STL POWER UP 

This question is done using map. We map the occurrence of a particular coin in the first side and the second side. The condition if(arr[i]!=arr[i+1]) prevents updation in cases where the coin has same faces on both sides, else we inrement the occurence of the color on both the faces by 1.

Let  m be half that is (n+1)/2

Now, a[arr[i]] is the occurrence of the particular color on the front side and b[arr[i]] is the occurence of the same color in the back side. If for any colour a[arr[i]]>=m , then the answer is 0.

and if and only iff(a[arr[i]+b[arr[i]]>=m)

answers is minimum of (ans, m-a[arr[i]])  ((note:: m - a[arr[i]] is the number of coins that are to be faced up)