C. Mr. Perfectly Fine

time limit per test

2 seconds

memory limit per test

256 megabytes

Victor wants to become "Mr. Perfectly Fine". For that, he needs to acquire a certain set of skills. More precisely, he has 22 skills he needs to acquire.

Victor has nn books. Reading book ii takes him mimi minutes and will give him some (possibly none) of the required two skills, represented by a binary string of length 22.

What is the minimum amount of time required so that Victor acquires all of the two skills?

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000) --- the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤2⋅1051≤n≤2⋅105) --- the number of books available.

Then nn lines follow. Line ii contains a positive integer mimi (1≤mi≤2⋅1051≤mi≤2⋅105) and a binary string of length 22, where si1=1si1=1 if reading book ii acquires Victor skill 11, and si1=0si1=0 otherwise, and si2=1si2=1 if reading book ii acquires Victor skill 22, and si2=0si2=0 otherwise.

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

For each test case, output a single integer denoting the minimum amount of minutes required for Victor to obtain both needed skills and −1−1 in case it's impossible to obtain the two skills after reading any amount of books.

Example

Input

Copy

复制代码

6

4

2 00

3 10

4 01

4 00

5

3 01

3 01

5 01

2 10

9 10

1

5 11

3

9 11

8 01

7 10

6

4 01

6 01

7 01

8 00

9 01

1 00

4

8 00

9 10

9 11

8 11

Output

Copy

复制代码
7
5
5
9
-1
8

Note

In the first test case, we can use books 22 and 33, with a total amount of minutes spent equal to 3+4=73+4=7.

In the second test case, we can use the books 11 and 44, with a total amount of minutes spent equal to 3+2=53+2=5.

In the third test case, we have only one option and that is reading book 11 for a total amount of minutes spent equal to 55.

解题说明:此题是一道模拟题,可以采用贪心算法,找出包含1和10的耗费最小时间组合或者是直接包含11的最小时间,两者进行比较即可。

cpp 复制代码
#include <stdio.h>

int main()
{
	int t = 0, n = 0, a = 0, s = 0;
	scanf("%d", &t);
	while (t--) 
	{
		scanf("%d", &n);
		int min_1 = 200005, min_2 = 200005, min_3 = 2000005;
		for (int i = 0; i < n; i++)
		{
			scanf("%d%d", &a, &s);
			if (1 == s && min_1 > a)
			{
				min_1 = a;
			}
			else if (10 == s && min_2 > a)
			{
				min_2 = a;
			}
			else if (11 == s && min_3 > a)
			{
				min_3 = a;
			}
		}
		if ((min_1 == 200005 || min_2 == 200005) && min_3 == 2000005)
		{
			printf("-1\n");
		}
		else if (min_1 + min_2 >= min_3)
		{
			printf("%d\n", min_3);
		}
		else
		{
			printf("%d\n", min_1 + min_2);
		}
	}
	return 0;
}
相关推荐
Ysjt | 深4 分钟前
C++多线程编程入门教程(优质版)
java·开发语言·jvm·c++
ephemerals__9 分钟前
【c++丨STL】list模拟实现(附源码)
开发语言·c++·list
码农飞飞13 分钟前
深入理解Rust的模式匹配
开发语言·后端·rust·模式匹配·解构·结构体和枚举
一个小坑货15 分钟前
Rust 的简介
开发语言·后端·rust
湫ccc22 分钟前
《Python基础》之基本数据类型
开发语言·python
Matlab精灵23 分钟前
Matlab函数中的隐马尔可夫模型
开发语言·matlab·统计学习
Microsoft Word24 分钟前
c++基础语法
开发语言·c++·算法
数据小爬虫@27 分钟前
如何利用java爬虫获得淘宝商品评论
java·开发语言·爬虫
qq_1728055934 分钟前
RUST学习教程-安装教程
开发语言·学习·rust·安装
wjs202442 分钟前
MongoDB 更新集合名
开发语言