Training is indispensable for achieving good results at ICPC. Rabbit wants to win at ICPC, so he decided to practice today as well.

Today's training is to gain dexterity that never mistypes by carefully stacking blocks. Since there are many building blocks, let's build a tall tower.

There are N blocks, and the i-th (1 ≤ i ≤ N) building blocks are in the shape of a rectangular parallelepiped of 1 x Ai x Bi. The side of length 1 is used in the depth direction, and the sides of lengths Ai and Bi are assigned one by one in the horizontal direction and one in the height direction. When building blocks, the upper building blocks must be exactly shorter in width than the lower building blocks. The blocks can be used in any order, and some blocks may not be used. Under these restrictions, I want to make the tallest tower that can be built.



Input


N
A1 B1
...
AN BN


Satisfy 1 ≤ N ≤ 1,000, 1 ≤ Ai, Bi ≤ 1,000,000. All input values ​​are integers.

Output

Output the maximum height of the tower on one line.

Examples

Input

3
10 40
10 40
20 30


Output

80


Input

4
1 2
2 3
3 4
4 1


Output

11
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
//#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

using Flow = ll;
using Cost = ll;
const ll MAX_V = 5000;
const Cost INF = std::numeric_limits<Cost>::max() / 8;

struct PrimalDual {
	struct Edge {
		ll d;
		Flow c, f;
		Cost w;
		ll r, is_r;
		Edge(ll d_, Flow c_, Flow f_, Cost w_, ll r_, bool is_r_)
		: d(d_), c(c_), f(f_), w(w_), r(r_), is_r(is_r_) {}
	};

	ll n;
	std::vector<std::vector<Edge> > g;
	PrimalDual(ll n_) : n(n_), g(std::vector<std::vector<Edge> >(n_)) {}

	void add_edge(ll src, ll dst, Flow cap, Cost cost) {  // 有向辺
		ll rsrc = g[dst].size();
		ll rdst = g[src].size();
		g[src].emplace_back(dst, cap, 0, cost, rsrc, false);
		g[dst].emplace_back(src, cap, cap, -cost, rdst, true);
	}

	Cost solve(ll s, ll t, Flow f) {
		vector<vector<Edge>> ng=g;
		Cost res = 0;
		static Cost h[MAX_V + 10], dist[MAX_V];
		static int prevv[MAX_V + 10], preve[MAX_V + 10];
		// std::vector<Cost> h(g.size()), dist(g.size());
		// std::vector<int> prevv(g.size()), preve(g.size());

		using pcv = std::pair<Cost, ll>;
		std::priority_queue<pcv, std::vector<pcv>, std::greater<pcv> > q;
		std::fill(h, h + n, 0);
		while (f > 0) {
			std::fill(dist, dist + n, INF);
			dist[s] = 0;
			q.emplace(0, s);
			while (q.size()) {
				Cost cd;
				ll v;
				std::tie(cd, v) = q.top();
				q.pop();
				if (dist[v] < cd) continue;
				for (ll i = 0; i < (ll)(ng[v].size()); ++i) {
					Edge &e = ng[v][i];
					if (residue(e) == 0) continue;
					if (dist[e.d] + h[e.d] > cd + h[v] + e.w) {
						dist[e.d] = dist[v] + e.w + h[v] - h[e.d];
						prevv[e.d] = v;
						preve[e.d] = i;
						q.emplace(dist[e.d], e.d);
					}
				}
			}

			if (dist[t] == INF) return INF;  // 経路が見つからなかった

			// s-t 間最短路に沿って目一杯流す
			for (ll i = 0; i < n; ++i) h[i] += dist[i];
			Flow d = f;
			for (ll v = t; v != s; v = prevv[v]) {
				d = std::min(d, residue(ng[prevv[v]][preve[v]]));
			}
			f -= d;
			res += d * h[t];
			for (ll v = t; v != s; v = prevv[v]) {
				Edge &e = ng[prevv[v]][preve[v]];
				e.f += d;
				ng[v][e.r].f -= d;
			}
		}
		return res;
	}

	Flow residue(const Edge &e) { return e.c - e.f; }

	void show() {
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < (int)(g[i].size()); ++j) {
				Edge &e = g[i][j];
				if (e.is_r) continue;
				printf("%3d->%3d (flow:%d)\n", i, (int)e.d, (int)e.f);
			}
		}
	}
};

ll n;

ll solve() {
	vector<ll> a(n);
	vector<ll> b(n);
	REP(i,n) cin>>a[i]>>b[i];
	vector<ll> num;
	REP(i,n) num.pb(a[i]);
	REP(i,n) num.pb(b[i]);
	UNIQUE(num);
	PrimalDual pd(n+(ll)num.size()+3);
	REP(i,n) pd.add_edge(0,1+i,1,0);
	REP(i,(ll)num.size()) pd.add_edge(1+n+i,n+(ll)num.size()+1,1,0);
    pd.add_edge(n+(ll)num.size()+2,n+(ll)num.size()+1,INF,0);
	REP(i,n) {
		ll idxa=(lower_bound(ALL(num),a[i])-num.begin());
		ll idxb=(lower_bound(ALL(num),b[i])-num.begin());
		pd.add_edge(1+i,1+n+idxa,1,-b[i]);
		pd.add_edge(1+i,1+n+idxb,1,-a[i]);
        pd.add_edge(1+i,n+(ll)num.size()+2,1,0);
	}
    return -pd.solve(0,n+(ll)num.size()+1,n);
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin>>n;
	cout<<solve()<<endl;
	return 0;
}


