[Solution] Directional Increase Codeforces Solution

[Solution] Directional Increase Codeforces Solution 

C. Directional Increase

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

We have an array of length n$n$. Initially, each element is equal to 0$0$ and there is a pointer located on the first element.

We can do the following two kinds of operations any number of times (possibly zero) in any order:

1. If the pointer is not on the last element, increase the element the pointer is currently on by 1$1$. Then move it to the next element.
2. If the pointer is not on the first element, decrease the element the pointer is currently on by 1$1$. Then move it to the previous element.

But there is one additional rule. After we are done, the pointer has to be on the first element.

Solution Click Below:-  CLICK HERE

You are given an array a$a$. Determine whether it's possible to obtain a$a$ after some operations or not.

Input

The first line contains a single integer t$t$ (1≤t≤1000)$(1\le t\le 1000)$  — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n$n$ (1≤n≤2⋅105)$(1\le n\le 2\cdot {10}^5)$  — the size of array a$a$.

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The second line of each test case contains n$n$ integers a1,a2,…,an$a_1,a_2,\dots ,a_n$ (−109≤ai≤109$-{10}^9\le a_i\le {10}^9$) — elements of the array.

It is guaranteed that the sum of n$n$ over all test cases doesn't exceed 2⋅105$2\cdot {10}^5$.

Output

For each test case, print "Yes" (without quotes) if it's possible to obtain a$a$ after some operations, and "No" (without quotes) otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).