/*
* LMIO 2019 Ptatoes and Fertilizers
* a - b shows the number of fertilizers spared
* Take the prefix sum of a - b
* The problem becomes:
- An operation allows increment
or decrement an element by 1
- Special case on the leftmost element
> can't be decremented, must be non-negative
- Special case on the rightmost element
> can't be incremented
- We want to make the array non-decreasing
* This can be solved with slope trick
* A similar problem: https://codeforces.com/contest/713/problem/C
* More about slope trick: https://codeforces.com/blog/entry/77298
*/
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <iostream>
#include <iomanip>
#include <queue>
using namespace std;
#define ll long long
int n;
ll pref[500005], a, b;
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n;
for(int i = 0; i < n; i ++){
cin >> a >> b;
pref[i] = a - b;
if(i > 0)
pref[i] += pref[i - 1];
}
ll ans = 0ll;
priority_queue<ll> pq;
pq.push(0ll);
for(int i = 0; i < n; i ++){
if(pref[i] < 0){
ans += 0ll - pref[i];
pref[i] = 0ll;
}else if(pref[i] > pref[n - 1]){
ans += pref[i] - pref[n - 1];
pref[i] = pref[n - 1];
}
pq.push(pref[i]);
pq.push(pref[i]);
ans += pq.top() - pref[i];
pq.pop();
}
cout << ans << endl;
return 0;
}
