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.
//#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define REP(u) for(int i=p[u];i!=-1;i=e[i].nxt)
#define PII pair<int,int>
#define ll long long
#define pb push_back
#define PLL pair<ll,ll>
#define fi first
#define se second
#define sz(x) (int)(x.size())
#define rd (rand()<<16^rand())
#define db double
#define gc getchar()//(_p1==_p2&&(_p2=(_p1=_buf)+fread(_buf,1,100000,stdin),_p1==_p2)?EOF:*_p1++)
using namespace std;
char _buf[100000],*_p1=_buf,*_p2=_buf;
inline int gi()
{
	int x=0,f=1;
	char ch=gc;
	while(ch<'0'||ch>'9')
	{
		if(ch=='-')f=-1;
		ch=gc;
	}
	while(ch>='0'&&ch<='9')
	{
		x=(x<<3)+(x<<1)+(ch^48);
		ch=gc;
	}
	return (f==1)?x:-x;
}
inline int max(int a,int b){return a>b?a:b;}
inline int min(int a,int b){return a<b?a:b;}
const int maxn=1e5+5;
inline void init()
{

}
vector<pair<int,int> >v;
int n;
ll ans=0;
inline void input()
{
	n=gi();
	FOR(i,1,n)
	{
		int a=gi(),c=gi();
		v.pb({a,c});ans+=c;
	}
}
inline void solve()
{
	sort(v.begin(),v.end());
	ll cur=v[0].first+v[0].second;
	FOR(i,1,sz(v)-1)
	{
		ll now=v[i].first+v[i].second;
		if(v[i].first>cur)ans+=v[i].first-cur;
		cur=max(cur,now);
	}
	printf("%lld\n",ans);
}
int main()
{
	int T=1;
	//T=gi();
	while(T--)
	{
		init();
		input();
		solve();
	}
	return 0;
}




