// #pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define ios ios::sync_with_stdio(0),cin.tie(nullptr),cout.tie(nullptr)
#define inf 0x3f3f3f3f
#define ll long long
#define endl '\n'
#define T() int ____t;cin>>____t;while(____t--)
using namespace std;
constexpr int N=2e6+10,mod=1e9+7;
#define int ll

template <typename T> // 选用int || long long
struct Mid_set
{
	multiset<T, greater<T>> st1; //小于中位数的数的集合
	multiset<T> st2; //大于等于中位数的集合 
	//   st2.size() ≥ st1.size();
	ll sum1 = 0, sum2 = 0;
	void split()
	{
		while (st1.size() < st2.size())
		{
			T now = *st2.begin();
			sum2 -= now;
			sum1 += now;
			st1.insert(now);
			st2.erase(st2.begin());
		}
		while (st1.size() > st2.size())
		{
			T now = *st1.begin();
			st2.insert(now);
			sum1 -= now;
			sum2 += now;
			st1.erase(st1.begin());
		}
	}
	void insert(T x)
	{
		st2.insert(x);
		sum2 += x;
		split();
	}
	void erase(T x)
	{
		if (st1.count(x))
		{
			st1.erase(st1.lower_bound(x));
			sum1 -= x;
		}
		else
		{
			st2.erase(st2.lower_bound(x));
			sum2 -= x;
		}
		split();
	}
	bool empty()
	{
		return st2.empty();
	}
	T query()//返回中位数
	{
		return *st2.begin();//有奇数个数
	}
	ll become_mid() // 通过±1,将所有数变相同所需最少操作数
	{
		T mid = query();
		int s1 = st1.size(), s2 = st2.size();
		ll ans = 1ll * s1 * mid - sum1 + sum2 - 1ll * s2 * mid;
		return ans;
	}
};
void solve(){
	int n;
	cin>>n;
	vector<int>a(n+1),b(n+1);
	Mid_set<int>p;
	for(int i=1;i<=n;i++){
		cin>>a[i]>>b[i];
		p.insert(a[i]);
	}
	int med=p.query();
	int l=1,r=1;
	int ans=med;
	for(int i=1;i<=n;i++){
		p.insert(b[i]);
		p.erase(a[i]);
		int now=p.query();
		ans=max(ans,now);
		r=i;
		if(now<med){
			for(int j=l;j<=r;j++){
				p.insert(a[j]);
				p.erase(b[j]);
			}
			l=i+1;
		}
	}
	cout<<ans<<endl;
}
signed main()
{
	ios;
//	T()
	solve();
	return 0;
}