#include <bits/stdc++.h>
using namespace std;

using ll = long long;

int main() {
    ios_base::sync_with_stdio(0); 
    cin.tie(0);

    int n; // n <=2000
    cin >> n;
    vector<pair<int,int>> v(n+1); //端子

    for(int i=1;i<=n;i++){
        cin >> v[i].first >> v[i].second;
    }
    sort(v.begin()+1,v.end(),[](auto &left, auto&right){
        return left.first > right.first;
    });

    
    //for(int i=1;i<=n;i++){
        //cout << v[i].first << v[i].second << endl;
    //}


    int dp[2001][2001];
    //vector<vector<int>> dp(n+1,vector<int>(n+1));

    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++){
        dp[0][i] = -1e9;
        //dp[i][n+1] = -1e9;
    }
    int ans = 0;
    dp[0][1] = 0;

    for(int i=1;i<=n;i++){
        for(int j=0;j<=n;j++){
            // 3 5 0 4 37
            dp[i][j] = max(dp[i-1][j], dp[i-1][min(n,max(j-v[i].first, 0) + 1)] + v[i].second);
            //cout << i << " " << j << " " << v[i].first << " " << v[i].second << " " << dp[i][j] << endl;
            ans = max(ans, dp[i][j]);
        }
    }

    cout << ans << "\n";



    // base case: dp[0][1] = 0
    // case 1: x take : dp[i][j] = dp[i-1][j]
    // case 2: take : dp[i][j] = dp[i-1][j- ai + 1] (ji-1 + ai - 1 = j)  + bi
    // take note of negative, actually if it negative we would make it one cuz er
    // if j - ai + 1 < 0, then make it one bcs 
    // if ai > j, the add ai =? 

    // それぞれのストラップは,携帯電話に直接取り付けるか,あるいは他のストラップの端子に取り付けることができる


}