Let‘s consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the largest of the sums of numbers that appear on the paths starting from the top towards the base, so that:

• on each path the next number is located on the row below, more precisely either directly below or below and one place to the right;
• the number of rows is strictly positive, but less than 100
• all numbers are positive integers between O and 99.

Input

In the first line integer n - the number of test cases (equal to about 1000).
Then n test cases follow. Each test case starts with the number of lines which is followed by their content.

Output

For each test case write the determined value in a separate line.

Example

Input:
2
3
1
2 1
1 2 3
4 
1 
1 2 
4 1 2
2 3 1 1 

Output:
5
9

使用动态规划法解决本题。

和一题leetcode题一样，从底往上查找路径。

#include <vector>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <iostream>
using namespace std;

int triPath(vector<vector<int> > &tri)
{
	if (tri.empty()) return 0;
	vector<int> path(tri.back());
	for (int i = (int)tri.size() - 2; i >= 0 ; i--)//unsigned做减法会溢出！！！
	{
		for (int j = 0; j < (int)tri[i].size(); j++)
		{
			path[j] = max(path[j], path[j+1]) + tri[i][j];
		}
	}
	return path.front();
}

int SumsinATriangle()
{
	int T, n;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &n);
		vector<vector<int> > tri;
		for (int i = 1; i <= n; i++)
		{
			vector<int> tmp(i);
			for (int j = 0; j < i; j++)
			{
				scanf("%d", &tmp[j]);
			}
			tri.push_back(tmp);
		}
		printf("%d\n", triPath(tri));
	}
	return 0;
}