Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each
test case start with a positive integer N(1<=N<=1000) which
indicate the number of homework.. Then 2 lines follow. The first line
contains N integers that indicate the deadlines of the subjects, and the
next line contains N integers that indicate the reduced scores.

For each test case, you should output the smallest total reduced score, one line per test case.

#include<iostream>

#include<algorithm>

#include<string.h>

using namespace std;

struct node {										//t代表时间，a代表分值

	int t, a;

}s[1010];

int T, n, cnt;

bool used[1010];									//标记某个时间段是否用了

bool cmp(node x, node y) {

	if (x.a == y.a)

		return x.t<y.t;

	else

		return x.a>y.a;

}

int main() {

	cin >> T;

	while (T--) {

		cin >> n;

		for (int i = 1; i <= n; i++)

			cin >> s[i].t;

		for (int i = 1; i <= n; i++)

			cin >> s[i].a;

		sort(s + 1, s + n + 1, cmp);

		memset(used, false, sizeof(used));				//不要忘了初始化

		cnt = 0;

		for (int i = 1; i <= n; i++) {					//枚举，先把分大的安排好

			int ans = 1;

			for (int j = s[i].t; j>0; j--) {			//如果分大的当前时间已经被安排了，那就向前找

				if (!used[j]) {

					used[j] = true;

					ans = 0;

					break;

				}

			}

			if (ans)cnt += s[i].a;						//找不到，就只能加上

		}

		cout << cnt << "\n";

	}

	return 0;

}