#include <bits/stdc++.h>
using namespace std;
const int N = 400005;
pair<int, int> p[N];
struct BIT {
  vector<int> bit;
  int maxn;
  BIT(int n) {
    bit.resize(n + 2, 0);
    maxn = n + 1;
  }
  void update(int k, int val) {
    for (int i = k; i < maxn; i += (i & (-i))) {
      bit[i] += val;
    }
  }
  int get(int k) {
    int res = 0;
    for (int i = k; i > 0; i -= (i & (-i))) {
      res += bit[i];
    }
    return res;
  }
};
int mx[N][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    int n;
    cin >> n;
    set<int> s;
    map<int, int> mp;
    for (int i = 1; i <= n; i++) {
      cin >> p[i].first >> p[i].second;
      s.insert(p[i].first);
      s.insert(p[i].second);
    }
    int cur = 1;
    for (auto u : s) {
      mp[u] = cur++;
    }
    for (int i = 1; i <= n; i++) {
      p[i].first = mp[p[i].first];
      p[i].second = mp[p[i].second];
    }
    BIT bit(2 * n + 2);
    for (int i = 1; i <= n; i++) {
      bit.update(p[i].first, 1);
      bit.update(p[i].second, -1);
    }
    BIT bit2(2 * n + 2);
    BIT bit3(2 * n + 2);
    BIT bit1(2 * n + 2);
    for (int i = 1; i <= n; i++) {
      bit3.update(p[i].second, 1);
      bit3.update(p[i].second + 1, -1);
      bit1.update(p[i].first, 1);
      bit1.update(p[i].first + 1, -1);
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (bit.get(i) == 1 && bit3.get(i) > 0) {
        bit2.update(i, 1);
      }
    }
    int maxi = 0;
    for (int i = 1; i <= n; i++) {
      int u = bit2.get(p[i].second - 1) - bit2.get(p[i].first - 1);
      if (bit.get(p[i].second - 1) == 1 && bit1.get(p[i].second) == 0) u--;
      maxi = max(maxi, u);
    }
    int ans = 0;
    sort(p + 1, p + n + 1);
    int righ = 0;
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      if (p[i].first > righ) ans++;
      if (p[i].first <= righ) flag = 1;
      righ = max(righ, p[i].second);
    }
    if (flag)
      cout << ans + maxi << '\n';
    else
      cout << ans - 1 << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 = 400005;
pair<int, int> p[N];
struct BIT {
  vector<int> bit;
  int maxn;
  BIT(int n) {
    bit.resize(n + 2, 0);
    maxn = n + 1;
  }
  void update(int k, int val) {
    for (int i = k; i < maxn; i += (i & (-i))) {
      bit[i] += val;
    }
  }
  int get(int k) {
    int res = 0;
    for (int i = k; i > 0; i -= (i & (-i))) {
      res += bit[i];
    }
    return res;
  }
};
int mx[N][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    int n;
    cin >> n;
    set<int> s;
    map<int, int> mp;
    for (int i = 1; i <= n; i++) {
      cin >> p[i].first >> p[i].second;
      s.insert(p[i].first);
      s.insert(p[i].second);
    }
    int cur = 1;
    for (auto u : s) {
      mp[u] = cur++;
    }
    for (int i = 1; i <= n; i++) {
      p[i].first = mp[p[i].first];
      p[i].second = mp[p[i].second];
    }
    BIT bit(2 * n + 2);
    for (int i = 1; i <= n; i++) {
      bit.update(p[i].first, 1);
      bit.update(p[i].second, -1);
    }
    BIT bit2(2 * n + 2);
    BIT bit3(2 * n + 2);
    BIT bit1(2 * n + 2);
    for (int i = 1; i <= n; i++) {
      bit3.update(p[i].second, 1);
      bit3.update(p[i].second + 1, -1);
      bit1.update(p[i].first, 1);
      bit1.update(p[i].first + 1, -1);
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (bit.get(i) == 1 && bit3.get(i) > 0) {
        bit2.update(i, 1);
      }
    }
    int maxi = 0;
    for (int i = 1; i <= n; i++) {
      int u = bit2.get(p[i].second - 1) - bit2.get(p[i].first - 1);
      if (bit.get(p[i].second - 1) == 1 && bit1.get(p[i].second) == 0) u--;
      maxi = max(maxi, u);
    }
    int ans = 0;
    sort(p + 1, p + n + 1);
    int righ = 0;
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      if (p[i].first > righ) ans++;
      if (p[i].first <= righ) flag = 1;
      righ = max(righ, p[i].second);
    }
    if (flag)
      cout << ans + maxi << '\n';
    else
      cout << ans - 1 << '\n';
  }
  return 0;
}

```