// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define dd double
#define ld long double
#define sl(n) scanf("%lld", &n)
#define si(n) scanf("%d", &n)
#define sd(n) scanf("%lf", &n)
#define pll pair <ll, ll>
#define pii pair <int, int>
#define mp make_pair
#define pb push_back
#define all(v) v.begin(), v.end()
#define inf (1LL << 61)
#define loop(i, start, stop, inc) for(ll i = start; i <= stop; i += inc)
#define for1(i, stop) for(ll i = 1; i <= stop; ++i)
#define for0(i, stop) for(ll i = 0; i < stop; ++i)
#define rep1(i, start) for(ll i = start; i >= 1; --i)
#define rep0(i, start) for(ll i = (start-1); i >= 0; --i)
#define ms(n, i) memset(n, i, sizeof(n))
#define casep(n) printf("Case %lld:", ++n)
#define pn printf("\n")
#define pf printf
#define EL '\n'
#define fastio std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

// === DEBUG MACRO STARTS HERE === //
#ifndef ONLINE_JUDGE
#define DEBUG
#define SYS_COL system("COLOR")
#endif

int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; SYS_COL; \
                cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":" \
                <<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}

template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream,ostream>::value,
            Ostream&>::type operator<<(Ostream& os,  const Cont& v) {
	os<<"[";
	for(auto& x:v){os<<x<<", ";}
	return os<<"]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os,  const pair<Ts...>& p){
	return os<<"{"<<p.first<<", "<<p.second<<"}";
}
#else
#define dbg(x)
#endif
// === DEBUG MACRO ENDS HERE === //

#define ff first
#define ss second

const ll sz = 2e5 + 10;
ll ara[sz];

int main()
{
    ll t;
    cin >> t;

    while(t--) {
        ll n; sl(n);
        for1(i, n) sl(ara[i]);

        set <int> lst;
        lst.insert(ara[1]);
        bool ok = 1;

        for(ll i = 2; i <= n; i++) {
            if(ara[i] == ara[i-1])
                continue;

            if(ara[i] > ara[i-1]) {
                auto it = lst.upper_bound(ara[i-1]);
                if(it != lst.end() && *it < ara[i]) {
                    ok = 0;
                    break;
                }
            }
            else {
                auto it = lst.lower_bound(ara[i-1]);
                if(it != lst.begin() && *(--it) > ara[i]) {
                    ok = 0;
                    break;
                }
            }

            lst.insert(ara[i]);
        }

        pf("%s", (ok)? "YES\n" : "NO\n");
    }

    return 0;
}

### Prompt

In cpp, your task is to solve 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
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define dd double
#define ld long double
#define sl(n) scanf("%lld", &n)
#define si(n) scanf("%d", &n)
#define sd(n) scanf("%lf", &n)
#define pll pair <ll, ll>
#define pii pair <int, int>
#define mp make_pair
#define pb push_back
#define all(v) v.begin(), v.end()
#define inf (1LL << 61)
#define loop(i, start, stop, inc) for(ll i = start; i <= stop; i += inc)
#define for1(i, stop) for(ll i = 1; i <= stop; ++i)
#define for0(i, stop) for(ll i = 0; i < stop; ++i)
#define rep1(i, start) for(ll i = start; i >= 1; --i)
#define rep0(i, start) for(ll i = (start-1); i >= 0; --i)
#define ms(n, i) memset(n, i, sizeof(n))
#define casep(n) printf("Case %lld:", ++n)
#define pn printf("\n")
#define pf printf
#define EL '\n'
#define fastio std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

// === DEBUG MACRO STARTS HERE === //
#ifndef ONLINE_JUDGE
#define DEBUG
#define SYS_COL system("COLOR")
#endif

int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; SYS_COL; \
                cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":" \
                <<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}

template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream,ostream>::value,
            Ostream&>::type operator<<(Ostream& os,  const Cont& v) {
	os<<"[";
	for(auto& x:v){os<<x<<", ";}
	return os<<"]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os,  const pair<Ts...>& p){
	return os<<"{"<<p.first<<", "<<p.second<<"}";
}
#else
#define dbg(x)
#endif
// === DEBUG MACRO ENDS HERE === //

#define ff first
#define ss second

const ll sz = 2e5 + 10;
ll ara[sz];

int main()
{
    ll t;
    cin >> t;

    while(t--) {
        ll n; sl(n);
        for1(i, n) sl(ara[i]);

        set <int> lst;
        lst.insert(ara[1]);
        bool ok = 1;

        for(ll i = 2; i <= n; i++) {
            if(ara[i] == ara[i-1])
                continue;

            if(ara[i] > ara[i-1]) {
                auto it = lst.upper_bound(ara[i-1]);
                if(it != lst.end() && *it < ara[i]) {
                    ok = 0;
                    break;
                }
            }
            else {
                auto it = lst.lower_bound(ara[i-1]);
                if(it != lst.begin() && *(--it) > ara[i]) {
                    ok = 0;
                    break;
                }
            }

            lst.insert(ara[i]);
        }

        pf("%s", (ok)? "YES\n" : "NO\n");
    }

    return 0;
}

```