#include <iostream>

using namespace std;

int n;//n个申请

int get[100005];//接受的收益

int giveup[100005];//需要放弃的个数

int dp[100005];//到第i个申请时，此时可收益的最大值

int main() { cin>>n;

for(int i=1;i<=n;i++){
	cin>>get[i];
	cin>>giveup[i];	
}

dp[1]=get[1];

for(int i=1;i<=n;i++){
	
	dp[i]=max(dp[i],dp[i-1]);
	
	if(i+giveup[i]<=n)//接受第i+giveup个，且不越界 
	{
	int rec=dp[i]+get[i];//新的dp[i+giveup[i]]，用rec记录			
	dp[i+giveup[i]]=max(dp[i+giveup[i]],rec); 
	
	}
	 
} 

cout<<dp[n];


return 0;

}

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int inf=1e7+10;
int n,dp[inf],a[inf],b[inf];
signed main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i]>>b[i];
	for(int i=n;i>=1;i--){
		dp[i]=max(dp[i+1],dp[i+b[i]+1]+a[i]);
		//cout<<i<<":"<<dp[i]<<endl; 
	}
	cout<<dp[1];
	return 0;
}

n = int(input())
ls = [0]*(n+5)
for i in range(n):
    a,b = map(int,input().split())
    ls[i] = [a,b]

dp = [0]*(n+5)

for i in range(n-1,-1,-1):
    dp[i] = dp[i+1]
    c = i + ls[i][1]+1
    dp[i] = max(dp[i],ls[i][0]+(dp[c] if c<n else 0 ))

print(max(dp))