You're given an array a_1, …, a_n of n non-negative integers.

Let's call it sharpened if and only if there exists an integer 1 ≤ k ≤ n such that a_1 < a_2 < … < a_k and a_k > a_{k+1} > … > a_n. 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 a_i>0 and assign a_i := a_i - 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 ⋅ 10^5).

The second line of each test case contains a sequence of n non-negative integers a_1, …, a_n (0 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

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.
var numRun= readline();
 
//zeroes are important and the number of units it is away from the end
//012343210 is an ideal array. as long as every digit is greater than the number it is from the end, it is ok
//3210 is ok; 0110 is not ok... need to find this example
//possible arrays could be 012345,543210,0123210
for(var n=0;n<numRun;n++){
    var k=readline();
    var s=readline().split(" ");
    var ans="No";
    if(checkUp(s)){ans="Yes";}
    if(checkDown(s)){ans="Yes";}
    if(checkMidOdd(s)){ans="Yes";}
    if(checkMidEven1(s)){ans="Yes";}
    if(checkMidEven2(s)){ans="Yes";}
    print(ans);
}//num runs


 //check left to right. Right-most can be 0 but then every other needs to be up 1
function checkUp(arr){
    for(var i=0;i<arr.length;i++){
        if(arr[i]>=i){}
        else{return false;}//one index is too low
    }//fpr
    return true;
}//checkUp

//right to left
function checkDown(arr){
    for(var i=0;i<arr.length;i++){
        if(arr[arr.length-1-i]>=i){}
        else{return false;}//one index is too low
    }//fpr
    return true;
}//checkUp

//check if it is a pyramid... highest index is in the middle
function checkMidOdd(arr){
        var mid=Math.floor(arr.length/2);
        for(var i=0;i<=mid;i++){
            if(arr[i]>=i){}
            else{return false;}//one index is too low
        }
        for(var i=0;i<mid;i++){
             if(arr[arr.length-1-i]>=i){}
             else{return false;}//one index is too low
        }
    return true;
}

//check if it is a pyramid... highest index is in the middle
//high point will be arr.length/2-1 (for 4 array, index 1)
function checkMidEven1(arr){
        if(arr.length%2===1){return false;}
        var mid=Math.floor(arr.length/2);
        if(arr[mid]<mid+1){return false;}
        for(var i=0;i<=mid;i++){
            if(arr[i]>=i){}
            else{return false;}//one index is too low
        }
        for(var i=0;i<mid;i++){
             if(arr[arr.length-1-i]>=i){}
             else{return false;}//one index is too low
        }
    return true;
}

function checkMidEven2(arr){
        if(arr.length%2===1){return false;}
        var mid=Math.floor(arr.length/2)-1;
        if(arr[mid]<mid+1){return false;}
        for(var i=0;i<=mid;i++){
            if(arr[i]>=i){}
            else{return false;}//one index is too low
        }
        for(var i=0;i<mid;i++){
             if(arr[arr.length-1-i]>=i){}
             else{return false;}//one index is too low
        }
    return true;
}