#include<bits/stdc++.h>
using namespace std;
 
#define int                   long long int
#define loop(i,a,b)           for(int i=(int)a;i<(int)b;++i)
#define loopr(i,a,b)          for(int i=(int)a;i>=(int)b;--i)
#define count_1(n)            __builtin_popcountll(n)
#define pb                    push_back
#define pc                    putchar
#define gc                    getchar
#define itoc(c)               ((char)(((int)'0')+c))
#define max(x,y)              ((x>y)?x:y)
#define min(x,y)              ((x<y)?x:y)
#define mid(s,e)              (s+(e-s)/2)
#define vi                    vector<int>
#define vvi                   vector<vector<int>>
#define pii                   pair<int,int>
const int MOD = 1000000007;
 
int modpow(int a, int b, int mod) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; res %= mod; a = a * a; a %= mod; }return res; }
int p(int a, int b) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; a = a * a; }return res; }
int gcd (int a, int b) { return b ? gcd (b, a % b) : a; }


void testcases(int test) {
    int n; cin>>n;
    vi a(n); loop(i,0,n) cin>>a[i];
    set<int> meme; meme.insert(a[0]);
    int flag = 1;
    loop(i,1,n) {
        if(a[i] == a[i-1]) continue;
        auto it1 = meme.begin(), it2 = meme.begin();
        if(a[i] < a[i-1]) { it1 = meme.upper_bound(a[i]); it2 = meme.lower_bound(a[i-1]); }
        else { it1 = meme.lower_bound(a[i]); it2 = meme.upper_bound(a[i-1]); }
        if(it1 != it2) { flag = 0; break; }
        meme.insert(a[i]);
    }
    if(flag) cout << "YES\n";
    else cout << "NO\n";
}     
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t; 
    cin >> t;
    // t = 1;
    loop(i,0,t) testcases(i);
}
### 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;
 
#define int                   long long int
#define loop(i,a,b)           for(int i=(int)a;i<(int)b;++i)
#define loopr(i,a,b)          for(int i=(int)a;i>=(int)b;--i)
#define count_1(n)            __builtin_popcountll(n)
#define pb                    push_back
#define pc                    putchar
#define gc                    getchar
#define itoc(c)               ((char)(((int)'0')+c))
#define max(x,y)              ((x>y)?x:y)
#define min(x,y)              ((x<y)?x:y)
#define mid(s,e)              (s+(e-s)/2)
#define vi                    vector<int>
#define vvi                   vector<vector<int>>
#define pii                   pair<int,int>
const int MOD = 1000000007;
 
int modpow(int a, int b, int mod) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; res %= mod; a = a * a; a %= mod; }return res; }
int p(int a, int b) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; a = a * a; }return res; }
int gcd (int a, int b) { return b ? gcd (b, a % b) : a; }


void testcases(int test) {
    int n; cin>>n;
    vi a(n); loop(i,0,n) cin>>a[i];
    set<int> meme; meme.insert(a[0]);
    int flag = 1;
    loop(i,1,n) {
        if(a[i] == a[i-1]) continue;
        auto it1 = meme.begin(), it2 = meme.begin();
        if(a[i] < a[i-1]) { it1 = meme.upper_bound(a[i]); it2 = meme.lower_bound(a[i-1]); }
        else { it1 = meme.lower_bound(a[i]); it2 = meme.upper_bound(a[i-1]); }
        if(it1 != it2) { flag = 0; break; }
        meme.insert(a[i]);
    }
    if(flag) cout << "YES\n";
    else cout << "NO\n";
}     
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t; 
    cin >> t;
    // t = 1;
    loop(i,0,t) testcases(i);
}
```