#include <bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(), (x).end()
#define int long long
#define SZ(x) ((int)(x).size())
#define fill(x, value) memset(x, value, sizeof(x))
#define debug(x) cout << #x << " ------> " << x << "\n"
typedef pair<int, int> PII;

const int N = 1e5 + 5;

string solve() {
    int n;
    cin >> n;
    int arr[n];
    for (int &x : arr) cin >> x;
    int mid = arr[0], lmax = -1e10, rmin = 1e10;
    // deque<int> L, R;
    // L.push_back(lmax);
    // R.push_back(rmin);
    set<int> st = {lmax, mid, rmin};
    auto it = ++st.begin();
    for (int i = 1; i < n; i++) {
        if(arr[i] > *next(it) or arr[i] < *prev(it)) return "NO";
        st.insert(arr[i]);
        it = st.find(arr[i]);

        // if (arr[i] == mid) {
        //     // R.push_front(mid);
        // } else if (arr[i] > mid) {
        //     if (arr[i] > R.front()) {
        //         return "NO";
        //     } else if (arr[i] == R.front()) {
        //         if (R.size() > 1) R.pop_front();
        //     } else {
        //         L.push_back(mid);
        //     }
        // } else {
        //     if (arr[i] < L.back())
        //         return "NO";
        //     else if (arr[i] == L.back()) {
        //         if (L.size() > 1) L.pop_back();
        //     } else {
        //         R.push_front(mid);
        //     }
        // }
        // mid = arr[i];
        // cout << L.back() << ' ' << mid << ' ' << R.back() << '\n';
    }
    return "YES";
}

signed main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
#endif

    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++) {
        // solve();
        cout << solve() << '\n';
    }

    cerr << "\nTime: " << clock() << " ms";
}
### 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 all(x) (x).begin(), (x).end()
#define int long long
#define SZ(x) ((int)(x).size())
#define fill(x, value) memset(x, value, sizeof(x))
#define debug(x) cout << #x << " ------> " << x << "\n"
typedef pair<int, int> PII;

const int N = 1e5 + 5;

string solve() {
    int n;
    cin >> n;
    int arr[n];
    for (int &x : arr) cin >> x;
    int mid = arr[0], lmax = -1e10, rmin = 1e10;
    // deque<int> L, R;
    // L.push_back(lmax);
    // R.push_back(rmin);
    set<int> st = {lmax, mid, rmin};
    auto it = ++st.begin();
    for (int i = 1; i < n; i++) {
        if(arr[i] > *next(it) or arr[i] < *prev(it)) return "NO";
        st.insert(arr[i]);
        it = st.find(arr[i]);

        // if (arr[i] == mid) {
        //     // R.push_front(mid);
        // } else if (arr[i] > mid) {
        //     if (arr[i] > R.front()) {
        //         return "NO";
        //     } else if (arr[i] == R.front()) {
        //         if (R.size() > 1) R.pop_front();
        //     } else {
        //         L.push_back(mid);
        //     }
        // } else {
        //     if (arr[i] < L.back())
        //         return "NO";
        //     else if (arr[i] == L.back()) {
        //         if (L.size() > 1) L.pop_back();
        //     } else {
        //         R.push_front(mid);
        //     }
        // }
        // mid = arr[i];
        // cout << L.back() << ' ' << mid << ' ' << R.back() << '\n';
    }
    return "YES";
}

signed main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
#endif

    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++) {
        // solve();
        cout << solve() << '\n';
    }

    cerr << "\nTime: " << clock() << " ms";
}
```