#include<bits/stdc++.h>
#define lld long long
#define pb push_back
#define mk make_pair
#define MAX (lld)1e18+7
#define lim (lld)1e9+7
#define MAX2 (lld)2e5+9
#define ff first
#define ss second
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;

const lld mod=lim;

lld power(lld x, lld y, lld p)  
{   lld res = 1;     x = x % p; if (x == 0) return 0;  
  
    while (y > 0)  
    { if (y & 1) res = (res*x) % p;  y = y>>1; x = (x*x) % p;   }  
    return res;  } 


lld extend_gcd(lld a, lld b, lld& x, lld& y) {
    if (b == 0) { x = 1;y = 0;return a;}
    lld x1, y1;
    lld d = extend_gcd(b, a % b, x1, y1);
    x = y1; y = x1 - y1 * (a / b);return d;	}

lld modinv(lld a, lld p)
{return power(a,p-2,p);}

lld rowNum[]={-1,0,0,1};
lld colNum[]={0,-1,1,0};

int main()
{
	fastio	
	lld t;
	cin>>t;
	while(t--)
	{
		lld n;
		cin>>n;
		std::vector<lld> v(n);
		for(lld i=0;i<n;i++)
			cin>>v[i];

		set<lld>s;
		s.insert(MAX);
		s.insert(-MAX);
		lld flag = 0;
		for(lld i=0;i<n-1;i++)
		{
			if(v[i]==v[i+1])
				continue;
			lld mx = max(v[i],v[i+1]);
			lld mn = min(v[i],v[i+1]);
			s.insert(v[i]);
			s.insert(v[i+1]);
			auto it1 = s.upper_bound(mn);
			auto it2 = s.lower_bound(mx);
			it2--;
			if(*it1!=mx || *it2!=mn)
			{	
				flag = 1;
				break;
			}
			
		}

		if(flag)
			cout<<"NO"<<endl;
		else
			cout<<"YES"<<endl;
	}
}
### Prompt

Please formulate a cpp solution to the following problem:

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>
#define lld long long
#define pb push_back
#define mk make_pair
#define MAX (lld)1e18+7
#define lim (lld)1e9+7
#define MAX2 (lld)2e5+9
#define ff first
#define ss second
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;

const lld mod=lim;

lld power(lld x, lld y, lld p)  
{   lld res = 1;     x = x % p; if (x == 0) return 0;  
  
    while (y > 0)  
    { if (y & 1) res = (res*x) % p;  y = y>>1; x = (x*x) % p;   }  
    return res;  } 


lld extend_gcd(lld a, lld b, lld& x, lld& y) {
    if (b == 0) { x = 1;y = 0;return a;}
    lld x1, y1;
    lld d = extend_gcd(b, a % b, x1, y1);
    x = y1; y = x1 - y1 * (a / b);return d;	}

lld modinv(lld a, lld p)
{return power(a,p-2,p);}

lld rowNum[]={-1,0,0,1};
lld colNum[]={0,-1,1,0};

int main()
{
	fastio	
	lld t;
	cin>>t;
	while(t--)
	{
		lld n;
		cin>>n;
		std::vector<lld> v(n);
		for(lld i=0;i<n;i++)
			cin>>v[i];

		set<lld>s;
		s.insert(MAX);
		s.insert(-MAX);
		lld flag = 0;
		for(lld i=0;i<n-1;i++)
		{
			if(v[i]==v[i+1])
				continue;
			lld mx = max(v[i],v[i+1]);
			lld mn = min(v[i],v[i+1]);
			s.insert(v[i]);
			s.insert(v[i+1]);
			auto it1 = s.upper_bound(mn);
			auto it2 = s.lower_bound(mx);
			it2--;
			if(*it1!=mx || *it2!=mn)
			{	
				flag = 1;
				break;
			}
			
		}

		if(flag)
			cout<<"NO"<<endl;
		else
			cout<<"YES"<<endl;
	}
}
```