#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 2e5 + 10;
int n, a[MAXN], b[MAXN];
int prv(int x) {
  int i = lower_bound(b, b + n, x) - b;
  if (i == 0) return -1;
  return b[i - 1];
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int ans = 0;
  map<int, int> dp_first, dp_last, dp_any, first, last, cnt;
  for (int i = 0; i < n; i++) {
    if (!first.count(a[i])) {
      first[a[i]] = i;
    }
    last[a[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int old = dp_any[a[i]];
    dp_any[a[i]] = max(old, dp_last[prv(a[i])]) + 1;
    ++cnt[a[i]];
    if (i == first[a[i]]) {
      if (dp_last[prv(a[i])] > 0) {
        dp_first[a[i]] = dp_last[prv(a[i])];
      } else {
        dp_first[a[i]] = cnt[prv(a[i])];
      }
    }
    if (i == last[a[i]]) {
      dp_last[a[i]] = dp_first[a[i]] + cnt[a[i]];
    }
    ans = max(ans, dp_any[a[i]]);
  }
  map<int, vector<int>> pos;
  for (int i = 0; i < n; i++) {
    pos[a[i]].push_back(i);
  }
  cnt.clear();
  for (int i = 0; i < n; i++) {
    ++cnt[a[i]];
    int me =
        pos[a[i]].end() - lower_bound(pos[a[i]].begin(), pos[a[i]].end(), i);
    int you = cnt[prv(a[i])];
    ans = max(ans, me + you);
  }
  cout << n - ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  cin >> te;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

This is a hard version of the problem. In this version, the given array can contain equal elements and the constraints on n are greater than in the easy version of the problem.

You are given an array a of n integers (the given array can contain equal elements). You can perform the following operations on array elements:

  1. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the begin of the array; 
  2. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the end of the array. 



For example, if n = 5, a = [4, 7, 2, 2, 9], then the following sequence of operations can be performed: 

  * after performing the operation of the first type to the second element, the array a will become [7, 4, 2, 2, 9]; 
  * after performing the operation of the second type to the second element, the array a will become [7, 2, 2, 9, 4]. 



You can perform operations of any type any number of times in any order.

Find the minimum total number of operations of the first and second type that will make the a array sorted in non-decreasing order. In other words, what is the minimum number of operations must be performed so the array satisfies the inequalities a[1] ≤ a[2] ≤ … ≤ a[n].

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.

Each test case starts with a line containing an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the array a.

Then follow n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — an array that needs to be sorted by the given operations. The given array can contain equal elements.

The sum of n for all test cases in one test does not exceed 2 ⋅ 10^5.

Output

For each test case output one integer — the minimum total number of operations of the first and second type, which will make the array sorted in non-decreasing order.

Example

Input


9
5
4 7 2 2 9
5
3 5 8 1 7
5
1 2 2 4 5
2
0 1
3
0 1 0
4
0 1 0 0
4
0 1 0 1
4
0 1 0 2
20
16 15 1 10 0 14 0 10 3 9 2 5 4 5 17 9 10 20 0 9


Output


2
2
0
0
1
1
1
1
16

Note

In the first test case, you first need to move two 2, to the beginning of the array. Therefore, the desired sequence of operations: [4, 7, 2, 2, 9] → [2, 4, 7, 2, 9] → [2, 2, 4, 7, 9].

In the second test case, you need to move the 1 to the beginning of the array, and the 8 — to the end. Therefore, the desired sequence of operations: [3, 5, 8, 1, 7] → [1, 3, 5, 8, 7] → [1, 3, 5, 7, 8].

In the third test case, the array is already sorted.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 2e5 + 10;
int n, a[MAXN], b[MAXN];
int prv(int x) {
  int i = lower_bound(b, b + n, x) - b;
  if (i == 0) return -1;
  return b[i - 1];
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int ans = 0;
  map<int, int> dp_first, dp_last, dp_any, first, last, cnt;
  for (int i = 0; i < n; i++) {
    if (!first.count(a[i])) {
      first[a[i]] = i;
    }
    last[a[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int old = dp_any[a[i]];
    dp_any[a[i]] = max(old, dp_last[prv(a[i])]) + 1;
    ++cnt[a[i]];
    if (i == first[a[i]]) {
      if (dp_last[prv(a[i])] > 0) {
        dp_first[a[i]] = dp_last[prv(a[i])];
      } else {
        dp_first[a[i]] = cnt[prv(a[i])];
      }
    }
    if (i == last[a[i]]) {
      dp_last[a[i]] = dp_first[a[i]] + cnt[a[i]];
    }
    ans = max(ans, dp_any[a[i]]);
  }
  map<int, vector<int>> pos;
  for (int i = 0; i < n; i++) {
    pos[a[i]].push_back(i);
  }
  cnt.clear();
  for (int i = 0; i < n; i++) {
    ++cnt[a[i]];
    int me =
        pos[a[i]].end() - lower_bound(pos[a[i]].begin(), pos[a[i]].end(), i);
    int you = cnt[prv(a[i])];
    ans = max(ans, me + you);
  }
  cout << n - ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  cin >> te;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}

```