There are n cities numbered from 1 to n, and city i has beauty a_i.

A salesman wants to start at city 1, visit every city exactly once, and return to city 1.

For all i≠ j, a flight from city i to city j costs max(c_i,a_j-a_i) dollars, where c_i is the price floor enforced by city i. Note that there is no absolute value. Find the minimum total cost for the salesman to complete his trip.

Input

The first line contains a single integer n (2≤ n≤ 10^5) — the number of cities.

The i-th of the next n lines contains two integers a_i, c_i (0≤ a_i,c_i≤ 10^9) — the beauty and price floor of the i-th city.

Output

Output a single integer — the minimum total cost.

Examples

Input


3
1 9
2 1
4 1


Output


11


Input


6
4 2
8 4
3 0
2 3
7 1
0 1


Output


13

Note

In the first test case, we can travel in order 1→ 3→ 2→ 1. 

  * The flight 1→ 3 costs max(c_1,a_3-a_1)=max(9,4-1)=9. 
  * The flight 3→ 2 costs max(c_3, a_2-a_3)=max(1,2-4)=1. 
  * The flight 2→ 1 costs max(c_2,a_1-a_2)=max(1,1-2)=1. 



The total cost is 11, and we cannot do better.
#include<bits/stdc++.h>
#define rep(i,a,b) for(ll i=(a);i<=(b);i++)
#define per(i,a,b) for(ll i=(a);i>=(b);i--)
#define lc(x) (x<<1)
#define rc(x) (lc(x)|1)
#define lowbit(x) (x&-x)
#define Max(a,b) (a>b?a:b)
#define Min(a,b) (a<b?a:b)
#define next Cry_For_theMoon
#define il inline
#define pb(x) push_back(x)
#define is(x) insert(x)
#define sit set<int>::iterator
#define mapit map<int,int>::iterator
#define pi pair<int,int>
#define ppi pair<int,pi>
#define pp pair<pi,pi>
#define fr first
#define se second
#define vit vector<int>::iterator
#define mp(x,y) make_pair(x,y)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef double db;
using namespace std;
const int MAXN=2e5+10;
pi a[MAXN];
ll n,ans,idx,pre[MAXN]; 
int LowerBound(int val){
	//最后一个小于等于val的 
	int L=1,R=n,ans=n;
	while(L<=R){
		int mid=(L+R)>>1;
		if(a[mid].fr<=val){
			ans=mid;
			L=mid+1;
		}else{
			R=mid-1;
		}
	}
	return ans;
}
int main(){
	cin>>n;
	rep(i,1,n){
		cin>>a[i].fr>>a[i].se;ans+=a[i].se;
	} 
	sort(a+1,a+1+n);
	rep(i,1,n){
		pre[i]=Max(pre[i-1],a[i].fr+a[i].se);
	}
	idx=1;
	while(idx<n){
		int pos=LowerBound(a[idx].fr+a[idx].se);
		if(pos==n)break;
		//取[1,pos]最小值 
		ans+=Max(0,a[pos+1].fr-pre[pos]);
		idx=pos+1;
	}
	cout<<ans;
	return 0;
}
