You are given an easy task by your supervisor -- to find the best value of X, one of the parameters in an evaluation function, in order to improve the accuracy of the whole program.

However, after a few days‘ analysis, you realize that it is far harder than you imagine. There are so many values X can be, and the only way to find the best one among them is to try all these possible values one after another!

Fortunately, you know that X is an integer and thanks to the previous works by your senior fellow apprentices, you have got n constraints on X. Each constraint must be in one of the following forms:
1. < k: means that X is less than integer k;
2. > k: means that X is greater than integer k;
3. <= k: means that X is less than or equal to integer k;
4. >= k: means that X is greater than or equal to integer k;
5. = k: means that X is equal to integer k.

Now, you are going to figure out how many possible values X can be, so that you can estimate whether it is possible to finish your task before deadline.

The first line contains an integer T (1 ≤ T ≤ 10) -- the number of test cases.

For each test case:
The first line contains an integer n. 0 ≤ n ≤ 10 000.
Then follows n lines, each line contains a comparison operator o and an integer k, separated by a single space. o can be one of “>”, “<”, “>=”, “<=”, and “=”. 0 ≤ | k | ≤ 1 000 000 000.
There is no contradictory between these constraints, in other word, at least one integer value meets all of them.

For each test case, output one integer in a single line -- the number of possible values of X, or “-1” if the answer is infinite.

1
2
> 2
<= 5

3

#include<stdio.h>
#define ll long long
#define inf 9999999999
int main()
{
    ll t,n,a,l,r;
    char s[5];
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld",&n);
        l=-inf; r=inf;
        int flag=1;
        while(n--)
        {
            scanf("%s%lld",s,&a);
            if(s[1]!='\0'&&flag)
            {
                if(s[0]=='>')if(l<a)l=a;
                if(s[0]=='<'&&r>a)r=a;
            }
            else if(flag)
            {
                if(s[0]=='>'&&l<a+1)l=a+1;
                if(s[0]=='<'&&r>a-1)r=a-1;
                if(s[0]=='=')
                if(l<=a&&a<=r)l=r=a;else flag=0;
            }
        }
        if(flag==0||l>r)printf("0\n");
        else if(l==-inf||r==inf)printf("-1\n");
        else printf("%lld\n",r-l+1);

    }
}