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.
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define all(x)     		(x).begin(),(x).end()
#define ub    			upper_bound
#define lb 				lower_bound
#define ins				insert
#define int 			long long
#define endl			'\n'
#define ff              first
#define ss              second
#define ull				unsigned long long
#define pb              push_back
#define mp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define si 				set<int>
#define usi 			unordered_set<int>		
#define mii             map<int,int>
#define umii			unordered_map<int ,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<pii>
#define rep(i,a,n)		for(int i=a;i<n;++i)
#define repd(i,a,n)		for(int i=a;i>n;--i)
#define setbits(x)      __builtin_popcountll(x)
#define zerobits(x)      __builtin_ctzll(x)
#define Mod             1000000007
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
#define deb(x)			cout<<#x <<"="<<x<<endl;
#define deb2(x,y)		cout<<#x<<"="<<x<<"  "<<#y<<"="<<y<<endl;
mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
int Pow(int a,int b){int res=1;while(b>0){if(b&1)res*=a;a*=a;b/=2;}return res;}
int MPow(int a,int  b){int res=1;while(b>0){if(b&1)res=(res*a)%Mod;a=(a*a)%Mod;b/=2;}return res%Mod;}
int log(int x){return 64-__builtin_clzll(x)-1;}

void c_p_p()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
}
struct Node
{
	int a,b,c,d;
	
};
bool cmp(Node A,Node B)
{
	return A.a<B.a;
}
const int Max=1e5+1;

int32_t main()
{
	c_p_p();
	w(t)
	{
		int n;
		cin>>n;
		int A[n];
		rep(i,0,n)cin>>A[i];
		int ele=0,foo=1;
		rep(i,0,n)
		{
			if(A[i]<i)
			{
				ele=A[i-1];
				//deb(ele);
				int ind=0;
				repd(j,n-1,i-1)
				{
					if(A[i]<ind)
					{
						foo=0;

					}
					++ind;

				}
				//deb(i);
				if(ele<n-i)foo=0;
				break;
			}

		}
		if(foo)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}
	