#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 11;
const int inf = 2e9;
int dp[N];
int tree[4 * N];
void update(int low, int high, int idx, int val, int pos) {
  if (low > idx || high < idx) return;
  if (low == high) {
    tree[pos] = val;
    return;
  }
  int mid = (low + high) / 2;
  update(low, mid, idx, val, 2 * pos + 1);
  update(mid + 1, high, idx, val, 2 * pos + 2);
  tree[pos] = min(tree[2 * pos + 1], tree[2 * pos + 2]);
}
int query(int low, int high, int qlow, int qhigh, int pos) {
  if (low > qhigh || high < qlow) return inf;
  if (low >= qlow && high <= qhigh) return tree[pos];
  int mid = (low + high) / 2;
  return min(query(low, mid, qlow, qhigh, 2 * pos + 1),
             query(mid + 1, high, qlow, qhigh, 2 * pos + 2));
}
int32_t main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<int, int>> arr(n);
    for (int i = 0; i < n; ++i) cin >> arr[i].first >> arr[i].second;
    sort(arr.begin(), arr.end());
    int prev = inf;
    for (int i = n - 1; i >= 0; --i) {
      int f = arr[i].first;
      int s = arr[i].second;
      if (prev > s)
        dp[i] = 1 + dp[i + 1];
      else {
        int idx = upper_bound(arr.begin(), arr.end(), make_pair(s, inf)) -
                  arr.begin();
        int val = query(0, n - 1, i + 1, idx - 1, 0);
        dp[i] = val;
      }
      update(0, n - 1, i, dp[i], 0);
      prev = f;
    }
    int MAX = arr[0].second;
    int ans = dp[1];
    int cnt = 1;
    for (int i = 1; i < n; ++i) {
      int idx = upper_bound(arr.begin(), arr.end(), make_pair(MAX, inf)) -
                arr.begin();
      if (idx <= i + 1)
        ans = max(ans, cnt + dp[i + 1]);
      else {
        int val = query(0, n - 1, i + 1, idx - 1, 0);
        ans = max(ans, cnt + val - 1);
      }
      if (arr[i].first > MAX) cnt++;
      MAX = max(MAX, arr[i].second);
    }
    for (int i = 0; i < n; ++i) dp[i] = 0;
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

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

There are n segments on a Ox axis [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Segment [l, r] covers all points from l to r inclusive, so all x such that l ≤ x ≤ r.

Segments can be placed arbitrarily — be inside each other, coincide and so on. Segments can degenerate into points, that is l_i=r_i is possible.

Union of the set of segments is such a set of segments which covers exactly the same set of points as the original set. For example:

  * if n=3 and there are segments [3, 6], [100, 100], [5, 8] then their union is 2 segments: [3, 8] and [100, 100]; 
  * if n=5 and there are segments [1, 2], [2, 3], [4, 5], [4, 6], [6, 6] then their union is 2 segments: [1, 3] and [4, 6]. 



Obviously, a union is a set of pairwise non-intersecting segments.

You are asked to erase exactly one segment of the given n so that the number of segments in the union of the rest n-1 segments is maximum possible.

For example, if n=4 and there are segments [1, 4], [2, 3], [3, 6], [5, 7], then:

  * erasing the first segment will lead to [2, 3], [3, 6], [5, 7] remaining, which have 1 segment in their union; 
  * erasing the second segment will lead to [1, 4], [3, 6], [5, 7] remaining, which have 1 segment in their union; 
  * erasing the third segment will lead to [1, 4], [2, 3], [5, 7] remaining, which have 2 segments in their union; 
  * erasing the fourth segment will lead to [1, 4], [2, 3], [3, 6] remaining, which have 1 segment in their union. 



Thus, you are required to erase the third segment to get answer 2.

Write a program that will find the maximum number of segments in the union of n-1 segments if you erase any of the given n segments.

Note that if there are multiple equal segments in the given set, then you can erase only one of them anyway. So the set after erasing will have exactly n-1 segments.

Input

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

The first of each test case contains a single integer n (2 ≤ n ≤ 2⋅10^5) — the number of segments in the given set. Then n lines follow, each contains a description of a segment — a pair of integers l_i, r_i (-10^9 ≤ l_i ≤ r_i ≤ 10^9), where l_i and r_i are the coordinates of the left and right borders of the i-th segment, respectively.

The segments are given in an arbitrary order.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.

Output

Print t integers — the answers to the t given test cases in the order of input. The answer is the maximum number of segments in the union of n-1 segments if you erase any of the given n segments.

Example

Input


3
4
1 4
2 3
3 6
5 7
3
5 5
5 5
5 5
6
3 3
1 1
5 5
1 5
2 2
4 4


Output


2
1
5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 11;
const int inf = 2e9;
int dp[N];
int tree[4 * N];
void update(int low, int high, int idx, int val, int pos) {
  if (low > idx || high < idx) return;
  if (low == high) {
    tree[pos] = val;
    return;
  }
  int mid = (low + high) / 2;
  update(low, mid, idx, val, 2 * pos + 1);
  update(mid + 1, high, idx, val, 2 * pos + 2);
  tree[pos] = min(tree[2 * pos + 1], tree[2 * pos + 2]);
}
int query(int low, int high, int qlow, int qhigh, int pos) {
  if (low > qhigh || high < qlow) return inf;
  if (low >= qlow && high <= qhigh) return tree[pos];
  int mid = (low + high) / 2;
  return min(query(low, mid, qlow, qhigh, 2 * pos + 1),
             query(mid + 1, high, qlow, qhigh, 2 * pos + 2));
}
int32_t main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<int, int>> arr(n);
    for (int i = 0; i < n; ++i) cin >> arr[i].first >> arr[i].second;
    sort(arr.begin(), arr.end());
    int prev = inf;
    for (int i = n - 1; i >= 0; --i) {
      int f = arr[i].first;
      int s = arr[i].second;
      if (prev > s)
        dp[i] = 1 + dp[i + 1];
      else {
        int idx = upper_bound(arr.begin(), arr.end(), make_pair(s, inf)) -
                  arr.begin();
        int val = query(0, n - 1, i + 1, idx - 1, 0);
        dp[i] = val;
      }
      update(0, n - 1, i, dp[i], 0);
      prev = f;
    }
    int MAX = arr[0].second;
    int ans = dp[1];
    int cnt = 1;
    for (int i = 1; i < n; ++i) {
      int idx = upper_bound(arr.begin(), arr.end(), make_pair(MAX, inf)) -
                arr.begin();
      if (idx <= i + 1)
        ans = max(ans, cnt + dp[i + 1]);
      else {
        int val = query(0, n - 1, i + 1, idx - 1, 0);
        ans = max(ans, cnt + val - 1);
      }
      if (arr[i].first > MAX) cnt++;
      MAX = max(MAX, arr[i].second);
    }
    for (int i = 0; i < n; ++i) dp[i] = 0;
    cout << ans << '\n';
  }
  return 0;
}

```