### CSES Problem Set

You have to process $n$ tasks. Each task has a duration and a deadline, and you will process the tasks in some order one after another. Your reward for a task is $d-f$ where $d$ is its deadline and $f$ is your finishing time. (The starting time is $0$, and you have to process all tasks even if a task would yield negative reward.)

What is your maximum reward if you act optimally?

Input

The first input line has an integer $n$: the number of tasks.

After this, there are $n$ lines that describe the tasks. Each line has two integers $a$ and $d$: the duration and deadline of the task.

Output

Print one integer: the maximum reward.

Constraints
• $1\le n\le 2\cdot {10}^{5}$
• $1\le a,d\le {10}^{6}$
Example

Input:
36 108 155 12

Output:
2

#include <bits/stdc++.h>using namespace std;#define int            long long int#define F              first#define S              second#define pb             push_back#define si             set <int>#define vi             vector <int>#define pii            pair <int, int>#define vpi            vector <pii>#define vpp            vector <pair<int, pii>>#define mii            map <int, int>#define mpi            map <pii, int>#define spi            set <pii>#define endl           "\n"#define sz(x)          ((int) x.size())#define all(p)         p.begin(), p.end()#define double         long double#define que_max        priority_queue <int>#define que_min        priority_queue <int, vi, greater<int>>#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)#define print(a)       for(auto x : a) cout << x << " "; cout << endl#define print1(a)      for(auto x : a) cout << x.F << " " << x.S << endl#define print2(a,x,y)  for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endlinline int power(int a, int b){  int x = 1;  while (b)  {    if (b & 1) x *= a;    a *= a;    b >>= 1;  }  return x;}template <typename Arg1>void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }template <typename Arg1, typename... Args>void __f (const char* names, Arg1&& arg1, Args&&... args){  const char* comma = strchr (names + 1, ',');  cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);}const int N = 200005;bool cmp(pii &a, pii&b){    if(a.first!=b.first)return a.first<b.first;    return a.second<b.second;}void solve() {int n;cin>>n;vpi v(n);for (int i = 0; i < n; i++){    cin>>v[i].first;    cin>>v[i].second;}sort(v.begin(),v.end(),cmp);int ans=0;int t=0;for (int i = 0; i < n; i++){    t+=v[i].first;    ans+=v[i].second-t;}cout<<ans<<endl;}int32_t main(){  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);  clock_t z = clock();  int t = 1;//   cin >> t;   while (t--) solve();  return 0;}