We have a directed weighted graph with N vertices. Each vertex has two integers written on it, and the integers written on Vertex i are A_i and B_i.

In this graph, there is an edge from Vertex x to Vertex y for all pairs 1 \leq x,y \leq N, and its weight is {\rm min}(A_x,B_y).

We will consider a directed cycle in this graph that visits every vertex exactly once. Find the minimum total weight of the edges in such a cycle.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9
* 1 \leq B_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print the minimum total weight of the edges in such a cycle.

Examples

Input

3
1 5
4 2
6 3


Output

7


Input

4
1 5
2 6
3 7
4 8


Output

10


Input

6
19 92
64 64
78 48
57 33
73 6
95 73


Output

227
#include <bits/stdc++.h>

#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 2e5 + 10;

int n, m;
int A[N], B[N];

struct Num {
	int o, ty, w;

	bool operator < (const Num& y) const { return w < y.w; }
}num[N];

int deg[N];

int main() {

	scanf("%d", &n);
	For(i, 1, n) {
		scanf("%d%d", &A[i], &B[i]);
		num[++m] = (Num){i, 0, A[i]}, num[++m] = (Num){i, 1, B[i]};
	}
	sort(num + 1, num + m + 1);

	long long ans = 0;
	int c[2] = {0};
	For(i, 1, n) {
		++deg[num[i].o], ++c[num[i].ty];
		ans += num[i].w;
	}
	For(i, 1, n) if (!deg[i]) {
		printf("%lld\n", ans);
		return 0;
	}
	if (!c[0] || !c[1]) { printf("%lld\n", ans); return 0; }
	
	if (num[n].o == num[n + 1].o && c[num[n].ty] > 1) ans += min(num[n + 1].w - num[n - 1].w, num[n + 2].w - num[n].w);
	else ans += num[n + 1].w - num[n].w;

	printf("%lld\n", ans);

	return 0;
}