#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <map>
#include <cstdio>
#include <stack>
#include <set>
#include <climits>
#include <utility>
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast", "-funroll-all-loops", "-ffast-math")
#pragma GCC target("avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
#define endl '\n'
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define bug system("pause")
#define re(i, a, b) for (int i = a; i <= b; i++)
#define rre(i, a) for (int i = 0; i < a; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
const int N = 5e5 + 10;
const int M = 2e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

map<int, int> mp;
int a[N];
int t, n;

int main()
{
	IOS;
	cin >> t;
	while (t--)
	{
		mp.clear();
		cin >> n;
		for (int i = 0; i < n; i++)
			cin >> a[i];
		int l = -1e9 - 1, r = 1e9 + 1, mid = a[0];
		bool flag = 1;
		mp[mid] = 1;
		for (int i = 1; i < n; i++)
		{
			map<int, int>::iterator it = mp.find(mid);
			if (it == mp.begin())
				l = -1e9 - 1;
			else
				l = (--it)->first;
			it = mp.find(mid);
			if ((++it) == mp.end())
				r = 1e9 + 1;
			else
				r = it->first;
			mid = a[i];
			mp[mid] = 1;
			if (l > a[i] || a[i] > r)
			{
				flag = 0;
				break;
			}
		}
		cout << (flag ? "YES\n" : "NO\n");
	}
	return 0;
}
### Prompt

Generate 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 <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <map>
#include <cstdio>
#include <stack>
#include <set>
#include <climits>
#include <utility>
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast", "-funroll-all-loops", "-ffast-math")
#pragma GCC target("avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
#define endl '\n'
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define bug system("pause")
#define re(i, a, b) for (int i = a; i <= b; i++)
#define rre(i, a) for (int i = 0; i < a; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
const int N = 5e5 + 10;
const int M = 2e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

map<int, int> mp;
int a[N];
int t, n;

int main()
{
	IOS;
	cin >> t;
	while (t--)
	{
		mp.clear();
		cin >> n;
		for (int i = 0; i < n; i++)
			cin >> a[i];
		int l = -1e9 - 1, r = 1e9 + 1, mid = a[0];
		bool flag = 1;
		mp[mid] = 1;
		for (int i = 1; i < n; i++)
		{
			map<int, int>::iterator it = mp.find(mid);
			if (it == mp.begin())
				l = -1e9 - 1;
			else
				l = (--it)->first;
			it = mp.find(mid);
			if ((++it) == mp.end())
				r = 1e9 + 1;
			else
				r = it->first;
			mid = a[i];
			mp[mid] = 1;
			if (l > a[i] || a[i] > r)
			{
				flag = 0;
				break;
			}
		}
		cout << (flag ? "YES\n" : "NO\n");
	}
	return 0;
}
```