#include <bits/stdc++.h>

using namespace std;

const int mod = 1e9 + 7;
const int N = 5e6 + 132;
const long long WOW = 1e12;

int n;

long long a[N];
map <long long, long long> mp;
set <pair <long long, int> > q;

void solve() {
	q.clear();
	cin >> n;
	for(int i = 1; i <= n; i++) 
		cin >> a[i], mp[a[i]] = 1;
	/*
	mp[-WOW] = 1;
	mp[WOW] = 1;

	for(int i = 1; i <= n; i++)
		a[i] = mp[a[i]];
	long long big = mp[WOW];
	long long small = mp[-WOW];
	*/
	long long med = 0;
	bool ok = 1;
	for(int i = 1; i <= n; i++) {
		if(i == 1) {
			q.insert(make_pair(a[i], i));
			med = a[i];
		}else {
			if(a[i] < med) {
				set <pair <long long, int> > :: iterator it = q.lower_bound(make_pair(a[i] + 1, 0));
				if(it == q.end()) assert(0);
				if((*it).first != med) {
					ok = 0;
					break;
				}
				med = a[i];
				q.insert(make_pair(-WOW, i));
				q.insert(make_pair(a[i], i));
			} else if(a[i] > med) {
				set <pair <long long, int> > :: iterator it = q.lower_bound(make_pair(a[i], 0));
				if(it == q.begin()) {
//					cout << i << "AAA";
					assert(0);
				}
				it--;
				if((*it).first != med) {
					ok = 0;
					break;
				}
				med = a[i];
				q.insert(make_pair(WOW, i));
				q.insert(make_pair(a[i], i));
			}else {
				q.insert(make_pair(WOW, i));
				q.insert(make_pair(-WOW, i));
			}
		}
	}
	if(!ok) cout << "NO\n";
	else cout << "YES\n";
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);


	int T;
	cin >> T;
	while(T--) solve();
	
}
























### Prompt

Construct a cpp code solution to the problem outlined:

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;

const int mod = 1e9 + 7;
const int N = 5e6 + 132;
const long long WOW = 1e12;

int n;

long long a[N];
map <long long, long long> mp;
set <pair <long long, int> > q;

void solve() {
	q.clear();
	cin >> n;
	for(int i = 1; i <= n; i++) 
		cin >> a[i], mp[a[i]] = 1;
	/*
	mp[-WOW] = 1;
	mp[WOW] = 1;

	for(int i = 1; i <= n; i++)
		a[i] = mp[a[i]];
	long long big = mp[WOW];
	long long small = mp[-WOW];
	*/
	long long med = 0;
	bool ok = 1;
	for(int i = 1; i <= n; i++) {
		if(i == 1) {
			q.insert(make_pair(a[i], i));
			med = a[i];
		}else {
			if(a[i] < med) {
				set <pair <long long, int> > :: iterator it = q.lower_bound(make_pair(a[i] + 1, 0));
				if(it == q.end()) assert(0);
				if((*it).first != med) {
					ok = 0;
					break;
				}
				med = a[i];
				q.insert(make_pair(-WOW, i));
				q.insert(make_pair(a[i], i));
			} else if(a[i] > med) {
				set <pair <long long, int> > :: iterator it = q.lower_bound(make_pair(a[i], 0));
				if(it == q.begin()) {
//					cout << i << "AAA";
					assert(0);
				}
				it--;
				if((*it).first != med) {
					ok = 0;
					break;
				}
				med = a[i];
				q.insert(make_pair(WOW, i));
				q.insert(make_pair(a[i], i));
			}else {
				q.insert(make_pair(WOW, i));
				q.insert(make_pair(-WOW, i));
			}
		}
	}
	if(!ok) cout << "NO\n";
	else cout << "YES\n";
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);


	int T;
	cin >> T;
	while(T--) solve();
	
}
























```