#include<bits/stdc++.h>		
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pr;
typedef vector<pair<int,int> > vpr;
typedef vector<vi> vvi;
#define ff first
#define ss second
#define fort(i,m,n) for(int i=m;i<n;i++)
#define fora(arr,n) for(int i=0;i<n;i++) cin>>arr[i];
#define begud(k) cerr<<"\t-->"<<#k<<" = "<<k<<"\n";
#define endl "\n"
#define pb push_back
const int mod = 1e9 + 7;
ll power(ll a,ll b)
{
	if(b==0)
		return 1;
	ll res = power(a,b/2);
	ll ans = res*res;
	ans %= mod;
	if(b%2==0)
		return ans;
	return (a*ans)%mod;
}
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b)                
{
	return (a.second < b.second); 
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);	
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		vi b(n);
		fora(b,n);
		bool poss = true;
		set<int> ms;
		ms.insert(b[0]);
		int prev_median = b[0];
		int median;
		fort(i,1,n)
		{
			median = b[i];
			if(median == prev_median)
				continue;
			ms.insert(b[i]);
			if(median > prev_median)
			{
				auto it = ms.find(median);
				if(it != ms.begin())
				{
					it--;
					if(*it != prev_median)
					{
						// begud(*it);
						poss = false;
					}
				}
			}
			else
			{
				auto it = ms.find(median);
				it++;
				if(it != ms.end())
				{
					if(*it != prev_median)
					{
						// begud(*it);
						// begud(i);
						// begud(prev_median);
						poss = false;
					}
				}
			}
			prev_median = median;
		}
		if(poss)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}
    return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

Uh oh! Ray lost his array yet again! However, Omkar might be able to help because he thinks he has found the OmkArray of Ray's array. The OmkArray of an array a with elements a_1, a_2, …, a_{2k-1}, is the array b with elements b_1, b_2, …, b_{k} such that b_i is equal to the median of a_1, a_2, …, a_{2i-1} for all i. Omkar has found an array b of size n (1 ≤ n ≤ 2 ⋅ 10^5, -10^9 ≤ b_i ≤ 10^9). Given this array b, Ray wants to test Omkar's claim and see if b actually is an OmkArray of some array a. Can you help Ray?

The median of a set of numbers a_1, a_2, …, a_{2i-1} is the number c_{i} where c_{1}, c_{2}, …, c_{2i-1} represents a_1, a_2, …, a_{2i-1} sorted in nondecreasing order. 

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array b.

The second line contains n integers b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9) — the elements of b.

It is guaranteed the sum of n across all test cases does not exceed 2 ⋅ 10^5. 

Output

For each test case, output one line containing YES if there exists an array a such that b_i is the median of a_1, a_2, ..., a_{2i-1} for all i, and NO otherwise. The case of letters in YES and NO do not matter (so yEs and No will also be accepted).

Examples

Input


5
4
6 2 1 3
1
4
5
4 -8 5 6 -7
2
3 3
4
2 1 2 3


Output


NO
YES
NO
YES
YES


Input


5
8
-8 2 -6 -5 -4 3 3 2
7
1 1 3 1 0 -2 -1
7
6 12 8 6 2 6 10
6
5 1 2 3 6 7
5
1 3 4 3 0


Output


NO
YES
NO
NO
NO

Note

In the second case of the first sample, the array [4] will generate an OmkArray of [4], as the median of the first element is 4.

In the fourth case of the first sample, the array [3, 2, 5] will generate an OmkArray of [3, 3], as the median of 3 is 3 and the median of 2, 3, 5 is 3.

In the fifth case of the first sample, the array [2, 1, 0, 3, 4, 4, 3] will generate an OmkArray of [2, 1, 2, 3] as 

  * the median of 2 is 2 
  * the median of 0, 1, 2 is 1 
  * the median of 0, 1, 2, 3, 4 is 2 
  * and the median of 0, 1, 2, 3, 3, 4, 4 is 3. 



In the second case of the second sample, the array [1, 0, 4, 3, 5, -2, -2, -2, -4, -3, -4, -1, 5] will generate an OmkArray of [1, 1, 3, 1, 0, -2, -1], as 

  * the median of 1 is 1 
  * the median of 0, 1, 4 is 1 
  * the median of 0, 1, 3, 4, 5 is 3 
  * the median of -2, -2, 0, 1, 3, 4, 5 is 1 
  * the median of -4, -2, -2, -2, 0, 1, 3, 4, 5 is 0 
  * the median of -4, -4, -3, -2, -2, -2, 0, 1, 3, 4, 5 is -2 
  * and the median of -4, -4, -3, -2, -2, -2, -1, 0, 1, 3, 4, 5, 5 is -1 



For all cases where the answer is NO, it can be proven that it is impossible to find an array a such that b is the OmkArray of a.

### Response

```cpp
#include<bits/stdc++.h>		
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pr;
typedef vector<pair<int,int> > vpr;
typedef vector<vi> vvi;
#define ff first
#define ss second
#define fort(i,m,n) for(int i=m;i<n;i++)
#define fora(arr,n) for(int i=0;i<n;i++) cin>>arr[i];
#define begud(k) cerr<<"\t-->"<<#k<<" = "<<k<<"\n";
#define endl "\n"
#define pb push_back
const int mod = 1e9 + 7;
ll power(ll a,ll b)
{
	if(b==0)
		return 1;
	ll res = power(a,b/2);
	ll ans = res*res;
	ans %= mod;
	if(b%2==0)
		return ans;
	return (a*ans)%mod;
}
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b)                
{
	return (a.second < b.second); 
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);	
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		vi b(n);
		fora(b,n);
		bool poss = true;
		set<int> ms;
		ms.insert(b[0]);
		int prev_median = b[0];
		int median;
		fort(i,1,n)
		{
			median = b[i];
			if(median == prev_median)
				continue;
			ms.insert(b[i]);
			if(median > prev_median)
			{
				auto it = ms.find(median);
				if(it != ms.begin())
				{
					it--;
					if(*it != prev_median)
					{
						// begud(*it);
						poss = false;
					}
				}
			}
			else
			{
				auto it = ms.find(median);
				it++;
				if(it != ms.end())
				{
					if(*it != prev_median)
					{
						// begud(*it);
						// begud(i);
						// begud(prev_median);
						poss = false;
					}
				}
			}
			prev_median = median;
		}
		if(poss)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}
    return 0;
}
```