A. Shifting Stacks

原题链接:

添加链接描述

题意:

You have n stacks of blocks. The i-th stack contains hi blocks and it’s height is the number of blocks in it. In one move you can take a block from the i-th stack (if there is at least one block) and put it to the i+1-th stack. Can you make the sequence of heights strictly increasing?

Note that the number of stacks always remains n: stacks don’t disappear when they have 0 blocks.

Input

First line contains a single integer t (1≤t≤104) — the number of test cases.

The first line of each test case contains a single integer n (1≤n≤100). The second line of each test case contains n integers hi (0≤hi≤109) — starting heights of the stacks.

It’s guaranteed that the sum of all n does not exceed 104.

Output

For each test case output YES if you can make the sequence of heights strictly increasing and NO otherwise.

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

input

6
2
1 2
2
1 0
3
4 4 4
2
0 0
3
0 1 0
4
1000000000 1000000000 1000000000 1000000000

output

YES
YES
YES
NO
NO
YES

Note

In the first test case there is no need to make any moves, the sequence of heights is already increasing.

In the second test case we need to move one block from the first stack to the second. Then the heights become 0 1.

In the third test case we could move one block from the first stack to the second and then from the second to the third, which would make the heights 3 4 5.

In the fourth test case we can’t make a move, but the sequence is not increasing, so the answer is NO.

In the fifth test case we can only make one move (from the second to the third stack), which would make the heights 0 0 1. Both 0 1 0 and 0 0 1 are not increasing sequences, so the answer is NO.

题解：

本题是让找出在所有的石头堆是否可以排成一个长度为n的一个递增的堆列，
我们可以每一次都将前面比i多出来的石头拿出来放到后面，每次判断上次多出来的加上它本来的石头是否>i,如果小于就输出“NO”，否则的话输出“YES”。

代码：

#include<iostream>
using namespace std;int a[105];
int main()
{int t;cin>>t;while(t--){int n;cin>>n;for(int i=0;i<n;i++)cin>>a[i];long long int dp=0;int f=0;for(int i=0;i<n;i++){if(dp+a[i]<i){cout<<"NO"<<endl;f=1;break;}dp+=(a[i]-i);a[i]-=i;}if(f==0)cout<<"YES"<<endl;}
}