/*
Terrible Sets
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 5448 Accepted: 2778
Description


紧贴 x 轴有一些互相挨着的矩形，给定每个矩形的长宽，求他们可以形成的最大矩形面积 
Let N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N, and w0 = 0. 
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑0<=j<=i-1wj <= x <= ∑0<=j<=iwj} 
Again, define set S = {A| A = WH for some W , H ∈ R+ and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}. 
Your mission now. What is Max(S)? 
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy. 
But for this one, believe me, it's difficult.
Input


The input consists of several test cases. For each case, n is given in a single line, and then followed by n lines, each containing wi and hi separated by a single space. The last line of the input is an single integer -1, indicating the end of input. You may assume that 1 <= n <= 50000 and w1h1+w2h2+...+wnhn < 109.
Output


Simply output Max(S) in a single line for each case.
Sample Input


3
1 2
3 4
1 2
3
3 4
1 2
3 4
-1
Sample Output


12
14*/


#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
struct rec
{
int w;
int h;
} data;


int main()
{
int n,ans,i,lasth,totalw,curarea;
while(scanf("%d",&n)&&n!=-1){
ans=0;
stack<rec> s;
lasth=0;
for(i=0;i<n;++i){
scanf("%d%d",&data.w,&data.h);
if(data.h>=lasth){
s.push(data);
}else{
totalw=0;
curarea=0;
while(!s.empty()&&s.top().h>data.h){
totalw+=s.top().w;
curarea=totalw*s.top().h;
if(curarea>ans)
ans=curarea;
s.pop();
}
//退栈之前压入一个高度为data.h,宽度为totalw+data.w的矩形 
totalw+=data.w;
data.w=totalw;
s.push(data);
}
lasth=data.h;
}
totalw=0;
curarea=0;
while(!s.empty()){
totalw+=s.top().w;
curarea=totalw*s.top().h;
if(curarea>ans) 
ans=curarea;
s.pop();
}
printf("%d\n",ans);
}
  return 0;
}