B. Array Sharpening

time limit per test1 second
memory limit per test256 megabytes

You're given an array a1,…,an of n non-negative integers.

Let's call it sharpened if and only if there exists an integer 1≤k≤n such that a1<a2<…ak+1>…>an. In particular, any strictly increasing or strictly decreasing array is sharpened. For example:

The arrays [4], [0,1], [12,10,8] and [3,11,15,9,7,4] are sharpened;
The arrays [2,8,2,8,6,5], [0,1,1,0] and [2,5,6,9,8,8] are not sharpened.
You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any i (1≤i≤n) such that ai>0 and assign ai:=ai−1.

Tell if it's possible to make the given array sharpened using some number (possibly zero) of these operations.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤t≤15 000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤3⋅105).

The second line of each test case contains a sequence of n non-negative integers a1,…,an (0≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.

Output

For each test case, output a single line containing "Yes" (without quotes) if it's possible to make the given array sharpened using the described operations, or "No" (without quotes) otherwise.

Example

input
10
1
248618
3
12 10 8
6
100 11 15 9 7 8
4
0 1 1 0
2
0 0
2
0 1
2
1 0
2
1 1
3
0 1 0
3
1 0 1
output
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No

Note

In the first and the second test case of the first test, the given array is already sharpened.

In the third test case of the first test, we can transform the array into [3,11,15,9,7,4] (decrease the first element 97 times and decrease the last element 4 times). It is sharpened because 3<11<15 and 15>9>7>4.

In the fourth test case of the first test, it's impossible to make the given array sharpened.

题意

给你一个长度为n的数组，你可以进行若干次操作，每次操作可以让一个数减一，但是最多使得这个数变成0.

现在问你在进行若干次操作后，问你能否找到一个k，满足a1<a2<...<ak, ak<ak+1<ak+2<....<an。

题解

视频题解：https://www.bilibili.com/video/av86529667/

贪心，我们考虑第i个数，如果他是在k的左边，那么我肯定希望这个数变成i。

同理，如果他是在k的右边，我希望这个数变成n-i-1.

我扫一遍，维护前缀是否能够都能变成左边。扫一遍后缀，看是否都能够变成右边，最后枚举一下k就可以了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 300005;
int l[maxn],r[maxn],a[maxn];
void solve(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
        l[i]=0;
        r[i]=0;
    }
    for(int i=0;i<n;i++){
        if(a[i]>=i){
            l[i]=1;
            if(i>0&&l[i-1]==0){
                l[i]=0;
            }
        }
    }
    for(int i=n-1;i>=0;i--){
        if(a[i]>=n-1-i){
            r[i]=1;
            if(i<n-1&&r[i+1]==0){
                r[i]=0;
            }
        }
    }
    for(int i=0;i<n;i++){
        if(l[i]==1&&r[i]==1){
            cout<<"Yes"<<endl;
            return;
        }
    }
    cout<<"No"<<endl;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}