#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
cin.tie(NULL)->ios_base::sync_with_stdio(false);
int n; cin >> n;
vector<pair<ll, ll>> v(n+1); // pval = prefix sum val
for(int i=1; i<=n; i++){
ll a, b; cin >> a >> b; v[i] = {a, b};
}
sort(v.begin(), v.end());
vector<ll> pval(n+1, 0), sz(n+1, 0);
for(int i=1; i<=n; i++){
sz[i] = v[i].first; pval[i] = pval[i-1] + v[i].second;
}
/*
idea: work in ranges, if a_i and a_j was picked,
it makes sense to pick all the paintings with sizes inbetween i and j
so it's now essentially a prefix sum problem
find max(pval[j] - pval[i-1] - (sz[j] - sz[i]))
= (pval[j] - sz[j]) - (pval[i-1] - sz[i])
and if we iterate j, it now turns into "just keep the best (min) pval[i-1] - sz[i]"
*/
ll ans = -1, mni = 1e18;
for(int i=1; i<=n; i++){
mni = min(mni, pval[i-1] - sz[i]);
ans = max(ans, pval[i] - sz[i] - mni);
//cout << ans << ' ';
}
cout << ans;
return 0;
}
