#include <bits/stdc++.h>  
using namespace std;
typedef long long int ll;
typedef long double ld;
#define P 1000000007
#define rep(i,n) for(int i=0;i<n;++i)
#define re(i,a,n) for(i=a;i<=n;++i)
#define repr(i,a,n) for(i=a;i>=n;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ub(v,val) upper_bound(v.begin(),v.end(),val)
#define np(str) next_permutation(str.begin(),str.end())
#define lb(v,val) lower_bound(v.begin(),v.end(),val)
#define sortv(vec) sort(vec.begin(),vec.end())
#define rev(p) reverse(p.begin(),p.end());
#define mset(a,val) memset(a,val,sizeof(a));
ll MOD = 998244353;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 

void solve(){
   ll n;
   cin>>n;
   ll *a = new ll[n];
   rep(i,n){
       cin>>a[i];
   }
   int flag = 0;
   set <ll> s;
   s.insert(a[0]);

   for(int i = 1; i<n; i++){
       s.insert(a[i]);
       auto it1 = s.find( min (a[i], a[i-1]));
       auto it2 = s.find( max ( a[i], a[i-1]));

       if( it1 == it2 )
            continue;
       
       it1++;

       if( it1 == it2 )
             continue;
       flag = 1;
       break;
   }
   // ok we can use terniaery Op xD 
   flag ? cout<<"NO\n" : cout<<"YES\n";

}
int main()
{
    fast_cin();
    ll t;
    cin >> t;
    for(int it=1;it<=t;it++) {
        solve();
    }
    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 <bits/stdc++.h>  
using namespace std;
typedef long long int ll;
typedef long double ld;
#define P 1000000007
#define rep(i,n) for(int i=0;i<n;++i)
#define re(i,a,n) for(i=a;i<=n;++i)
#define repr(i,a,n) for(i=a;i>=n;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ub(v,val) upper_bound(v.begin(),v.end(),val)
#define np(str) next_permutation(str.begin(),str.end())
#define lb(v,val) lower_bound(v.begin(),v.end(),val)
#define sortv(vec) sort(vec.begin(),vec.end())
#define rev(p) reverse(p.begin(),p.end());
#define mset(a,val) memset(a,val,sizeof(a));
ll MOD = 998244353;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 

void solve(){
   ll n;
   cin>>n;
   ll *a = new ll[n];
   rep(i,n){
       cin>>a[i];
   }
   int flag = 0;
   set <ll> s;
   s.insert(a[0]);

   for(int i = 1; i<n; i++){
       s.insert(a[i]);
       auto it1 = s.find( min (a[i], a[i-1]));
       auto it2 = s.find( max ( a[i], a[i-1]));

       if( it1 == it2 )
            continue;
       
       it1++;

       if( it1 == it2 )
             continue;
       flag = 1;
       break;
   }
   // ok we can use terniaery Op xD 
   flag ? cout<<"NO\n" : cout<<"YES\n";

}
int main()
{
    fast_cin();
    ll t;
    cin >> t;
    for(int it=1;it<=t;it++) {
        solve();
    }
    return 0;
}
```