Algorithm


A. Equalize Prices Again
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs ai�� coins.

You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.

However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.

On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.

So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1q1001≤�≤100) — the number of queries. Then q queries follow.

The first line of the query contains one integer n (1n100)1≤�≤100) — the number of goods. The second line of the query contains n integers a1,a2,,an�1,�2,…,�� (1ai1071≤��≤107), where ai�� is the price of the i-th good.

Output

For each query, print the answer for it — the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.

Example
input
Copy
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
output
Copy
3
2
1

 



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int q, n, sum;

int main() {
#ifndef ONLINE_JUDGE
	freopen("in", "r", stdin);
#endif

	scanf("%d", &q);
	for(int i = 0; i < q; ++i) {
		sum = 0;

		scanf("%d", &n);
		for(int j = 0, tmp; j < n; ++j)
			scanf("%d", &tmp), sum += tmp;

		printf("%d\n", (sum + (n - 1)) / n);
	}

	return 0;
}
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Input

x
+
cmd
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1

Output

x
+
cmd
3
2
1
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Demonstration


Codeforcess Solution A. Equalize Prices Again-Solution in C, C++, Java, Python ,Codeforcess Solution

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