#include <bits/stdc++.h>
using namespace std;
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(long long &x, long long y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(long long &x, long long y) { return (y > x) ? x = y, 1 : 0; }
const int N = 200000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const long double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
pair<int, int> a[N];
int R[N], C[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = (0); i < (n); i++) {
      cin >> a[i].first >> a[i].second;
    }
    sort(a, a + n);
    R[0] = -inf;
    C[0] = 0;
    int cnt = 0, curR = -inf;
    for (int i = (0); i < (n - 1); i++) {
      if (curR < a[i].first) {
        cnt++;
      }
      setmax(curR, a[i].second);
      R[i + 1] = curR;
      C[i + 1] = cnt;
    }
    int ans = C[n - 1];
    deque<pair<int, int> > st;
    for (int i = (n - 1); i > (0); i--) {
      while ((int)st.size() > 0 && a[i].second >= st.front().first) {
        setmax(a[i].second, st.front().second);
        st.pop_front();
      }
      if ((int)st.size() > 0) {
        if (a[i].first >= st.front().first) {
          while (true) {
          }
        }
      }
      st.push_front(a[i]);
      int tmp = upper_bound(st.begin(), st.end(), make_pair(R[i - 1], inf)) -
                st.begin();
      setmax(ans, C[i - 1] + (int)st.size() - tmp);
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(long long &x, long long y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(long long &x, long long y) { return (y > x) ? x = y, 1 : 0; }
const int N = 200000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const long double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
pair<int, int> a[N];
int R[N], C[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = (0); i < (n); i++) {
      cin >> a[i].first >> a[i].second;
    }
    sort(a, a + n);
    R[0] = -inf;
    C[0] = 0;
    int cnt = 0, curR = -inf;
    for (int i = (0); i < (n - 1); i++) {
      if (curR < a[i].first) {
        cnt++;
      }
      setmax(curR, a[i].second);
      R[i + 1] = curR;
      C[i + 1] = cnt;
    }
    int ans = C[n - 1];
    deque<pair<int, int> > st;
    for (int i = (n - 1); i > (0); i--) {
      while ((int)st.size() > 0 && a[i].second >= st.front().first) {
        setmax(a[i].second, st.front().second);
        st.pop_front();
      }
      if ((int)st.size() > 0) {
        if (a[i].first >= st.front().first) {
          while (true) {
          }
        }
      }
      st.push_front(a[i]);
      int tmp = upper_bound(st.begin(), st.end(), make_pair(R[i - 1], inf)) -
                st.begin();
      setmax(ans, C[i - 1] + (int)st.size() - tmp);
    }
    cout << ans << "\n";
  }
  return 0;
}

```