#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    int n;
    scanf("%d", &n);
    int l[n], r[n];
    vector<int> v;
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &l[i], &r[i]);
      v.push_back(l[i]);
      v.push_back(r[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    unordered_map<int, int> m;
    int ct = 1;
    for (auto x : v) {
      m[x] = ct++;
    }
    int a[2 * n + 1];
    memset(a, 0, sizeof(a));
    vector<pair<int, int> > p;
    for (int i = 0; i < n; i++) {
      l[i] = m[l[i]];
      r[i] = m[r[i]];
      p.push_back({l[i], 1});
      p.push_back({r[i], -1});
    }
    sort(p.begin(), p.end(), cmp);
    int num1[2 * n + 1];
    memset(num1, 0, sizeof(num1));
    int ct0 = 0;
    int cursum = 0;
    int prev = 0;
    for (auto x : p) {
      int d = x.second;
      cursum += d;
      if (d == 1 && cursum == 2 && prev == -1) {
        num1[x.first]++;
      } else if (d == -1 && cursum == 0) {
        ct0++;
      }
      prev = d;
    }
    for (int i = 1; i <= 2 * n; i++) {
      num1[i] += num1[i - 1];
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      ans = max(ans, num1[r[i]] - num1[l[i]]);
    }
    printf("%d\n", min(n - 1, ans + ct0));
  }
}

### Prompt

Your challenge is to write a cpp solution to 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;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    int n;
    scanf("%d", &n);
    int l[n], r[n];
    vector<int> v;
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &l[i], &r[i]);
      v.push_back(l[i]);
      v.push_back(r[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    unordered_map<int, int> m;
    int ct = 1;
    for (auto x : v) {
      m[x] = ct++;
    }
    int a[2 * n + 1];
    memset(a, 0, sizeof(a));
    vector<pair<int, int> > p;
    for (int i = 0; i < n; i++) {
      l[i] = m[l[i]];
      r[i] = m[r[i]];
      p.push_back({l[i], 1});
      p.push_back({r[i], -1});
    }
    sort(p.begin(), p.end(), cmp);
    int num1[2 * n + 1];
    memset(num1, 0, sizeof(num1));
    int ct0 = 0;
    int cursum = 0;
    int prev = 0;
    for (auto x : p) {
      int d = x.second;
      cursum += d;
      if (d == 1 && cursum == 2 && prev == -1) {
        num1[x.first]++;
      } else if (d == -1 && cursum == 0) {
        ct0++;
      }
      prev = d;
    }
    for (int i = 1; i <= 2 * n; i++) {
      num1[i] += num1[i - 1];
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      ans = max(ans, num1[r[i]] - num1[l[i]]);
    }
    printf("%d\n", min(n - 1, ans + ct0));
  }
}

```