Parade （规律）

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Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.

There will be n columns participating in the parade, thei-th column consists of l_i soldiers, who start to march from left leg, andr_i soldiers, who start to march from right leg.

The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, andR is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal|L - R|.

No more than once you can choose one column and tellall the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one indexi and swap values l_i andr_i.

Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.

Input

The first line contains single integer n (1 ≤ n ≤ 10⁵) — the number of columns.

The next n lines contain the pairs of integersl_i andr_i (1 ≤ l_i, r_i ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.

Output

Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or0 if the maximum beauty is already reached.

Consider that columns are numbered from 1 to n in the order they are given in the input data.

If there are several answers, print any of them.

Example

Input

Output

Input

2
6 5
5 6

Output

Input

Output

Note

In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.

If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.

It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.

题意：

给出左右了两个数组，要求调换左右数组同一位置的的数字，使得数组和差点绝对值最大；

且只能调换一次，输出该数字在数组中的位置

思路：

求出初始每个数组的和sum1，sum2，每次调换一个位置寻找最大值，并把该位置记录

假设a为num1的数字，b为num2同一位置的数字，现在调换a，b

sum1= sum1-a+b; sum2=sum2-b+a;

| sum1 - sum2 | = | (sum1 - sum2 - 2*a + 2*b) | （| a | 为a 的绝对值）

代码：

#include<stdio.h>
#define MAX 100005
int num1[MAX],num2[MAX];//存放左右数字
int fab(int a){//求绝对值
    if(a<0) a=-a;
    return a;
}
int main(){
    int n;
    scanf("%d",&n);
    int i;
    int sum1=0,sum2=0;
    for(i=1;i<=n;i++){
        scanf("%d%d",&num1[i],&num2[i]);
        sum1+=num1[i];
        sum2+=num2[i];
    }
    int maxx=fab(sum1-sum2);
    int besti=0;
    for(i=1;i<=n;i++){
       int a,b;
        a=num1[i];
        b=num2[i];
       if(maxx<fab(sum1-sum2-2*a+2*b)){
          besti=i;
          maxx=fab(sum1-sum2-2*a+2*b);
       }
    }
    printf("%d\n",besti);
return 0;
}