A. Everyone Loves to Sleep

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vlad, like everyone else, loves to sleep very much.

Every day Vlad has to do nn things, each at a certain time. For each of these things, he has an alarm clock set, the ii-th of them is triggered on hihi hours mimi minutes every day (0≤hi<24,0≤mi<600≤hi<24,0≤mi<60). Vlad uses the 2424-hour time format, so after h=12,m=59h=12,m=59 comes h=13,m=0h=13,m=0 and after h=23,m=59h=23,m=59 comes h=0,m=0h=0,m=0.

This time Vlad went to bed at HH hours MM minutes (0≤H<24,0≤M<600≤H<24,0≤M<60) and asks you to answer: how much he will be able to sleep until the next alarm clock.

If any alarm clock rings at the time when he went to bed, then he will sleep for a period of time of length 00.

Input

The first line of input data contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases in the test.

The first line of the case contains three integers nn, HH and MM (1≤n≤10,0≤H<24,0≤M<601≤n≤10,0≤H<24,0≤M<60) — the number of alarms and the time Vlad went to bed.

The following nn lines contain two numbers each hihi and mimi (0≤hi<24,0≤mi<600≤hi<24,0≤mi<60) — the time of the ii alarm. It is acceptable that two or more alarms will trigger at the same time.

Numbers describing time do not contain leading zeros.

Output

Output tt lines, each containing the answer to the corresponding test case. As an answer, output two numbers  — the number of hours and minutes that Vlad will sleep, respectively. If any alarm clock rings at the time when he went to bed, the answer will be 0 0.

Example

input

Copy

3

1 6 13

8 0

3 6 0

12 30

14 45

6 0

2 23 35

20 15

10 30

output

Copy

1 47
0 0
10 55

解题说明：此题是一道模拟题，可以先换算成按分钟来统计时间，一天总共1440分钟，然后计算出每一个闹钟在一天中的分钟。同时记录睡觉所在的分钟，进行比较，得到一个最小的差值即可。

#include
int main()
 
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, H, M;
		scanf("%d%d%d", &n, &H, &M);
		int i, h, m, T, td = 0, k = 1440;
		T = H * 60 + M;
		for (i = 0; i < n; i++)
		{
			scanf("%d%d", &h, &m);
			td = h * 60 + m;
			if (td >= T)
			{
				td -= T;
			}
			else
			{
				td += 1440 - T;
			}
 
			if (td < k)
			{
				k = td;
			}
		}
		printf("%d %d\n", k / 60, k % 60);
	}
	return 0;
}