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.

題意

給你一個長度為n的數組，你可以進行若干次操作，每次操作可以讓一個數減一，但是最多使得這個數變成0.

現在問你在進行若干次操作后，問你能否找到一個k，滿足a1<a2<...<ak, ak<ak+1<ak+2<....<an。

題解

視頻題解：https://www.bilibili.com/video/av86529667/

貪心，我們考慮第i個數，如果他是在k的左邊，那么我肯定希望這個數變成i。

同理，如果他是在k的右邊，我希望這個數變成n-i-1.

我掃一遍，維護前綴是否能夠都能變成左邊。掃一遍后綴，看是否都能夠變成右邊，最后枚舉一下k就可以了。

代碼

#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;
}