#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
int t;
int tree[802000];
void add(int x, int num) {
  while (x <= 801000) {
    tree[x] += num;
    x += x & -x;
  }
}
int query(int x) {
  int ans = 0;
  while (x) {
    ans += tree[x];
    x -= x & -x;
  }
  return ans;
}
int c[802000], b[802000];
void solve() {
  vector<int> l, r, a;
  int x, y, n, mx = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x >> y;
    l.push_back(x);
    r.push_back(y);
    a.push_back(x);
    a.push_back(y);
  }
  sort(a.begin(), a.end());
  int p = unique(a.begin(), a.end()) - a.begin();
  for (auto& u : l) {
    u = lower_bound(a.begin(), a.begin() + p, u) - a.begin() + 1;
    u <<= 1;
  }
  for (auto& u : r) {
    u = lower_bound(a.begin(), a.begin() + p, u) - a.begin() + 1;
    u <<= 1;
  }
  for (int i = 0; i < n; i++) {
    add(l[i], 1);
    add(r[i] + 1, -1);
    mx = max(mx, r[i]);
  }
  mx++;
  for (int i = 1; i <= mx; i++) {
    b[i] = query(i);
  }
  int ans = 0, mmax = INT_MIN, flag0 = 0, cnt = 1;
  for (int i = 1; i <= mx; i++) {
    if (!b[i] && b[i - 1]) ans++;
  }
  for (int i = 1; i <= mx; i++) {
    if (b[i] > 1) c[i] = cnt;
    if (b[i] <= 1) {
      int p = i;
      flag0 = 0;
      while (b[p] <= 1 && p <= mx) {
        if (!b[p]) flag0 = 1;
        p++;
      }
      if (!flag0) cnt++;
      for (int j = i; j <= p; j++) c[j] = cnt;
      i = p;
    }
  }
  for (int i = 0; i < n; i++) {
    int tt = c[r[i]] - c[l[i]];
    if (c[r[i]] == c[l[i]] && b[l[i]] == 1 && b[r[i]] == 1) tt = -1;
    mmax = max(mmax, tt);
  }
  cout << ans + mmax << '\n';
  for (int i = 0; i < n; i++) {
    add(l[i], -1);
    add(r[i] + 1, 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Generate 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;
#pragma GCC optimize(2)
int t;
int tree[802000];
void add(int x, int num) {
  while (x <= 801000) {
    tree[x] += num;
    x += x & -x;
  }
}
int query(int x) {
  int ans = 0;
  while (x) {
    ans += tree[x];
    x -= x & -x;
  }
  return ans;
}
int c[802000], b[802000];
void solve() {
  vector<int> l, r, a;
  int x, y, n, mx = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x >> y;
    l.push_back(x);
    r.push_back(y);
    a.push_back(x);
    a.push_back(y);
  }
  sort(a.begin(), a.end());
  int p = unique(a.begin(), a.end()) - a.begin();
  for (auto& u : l) {
    u = lower_bound(a.begin(), a.begin() + p, u) - a.begin() + 1;
    u <<= 1;
  }
  for (auto& u : r) {
    u = lower_bound(a.begin(), a.begin() + p, u) - a.begin() + 1;
    u <<= 1;
  }
  for (int i = 0; i < n; i++) {
    add(l[i], 1);
    add(r[i] + 1, -1);
    mx = max(mx, r[i]);
  }
  mx++;
  for (int i = 1; i <= mx; i++) {
    b[i] = query(i);
  }
  int ans = 0, mmax = INT_MIN, flag0 = 0, cnt = 1;
  for (int i = 1; i <= mx; i++) {
    if (!b[i] && b[i - 1]) ans++;
  }
  for (int i = 1; i <= mx; i++) {
    if (b[i] > 1) c[i] = cnt;
    if (b[i] <= 1) {
      int p = i;
      flag0 = 0;
      while (b[p] <= 1 && p <= mx) {
        if (!b[p]) flag0 = 1;
        p++;
      }
      if (!flag0) cnt++;
      for (int j = i; j <= p; j++) c[j] = cnt;
      i = p;
    }
  }
  for (int i = 0; i < n; i++) {
    int tt = c[r[i]] - c[l[i]];
    if (c[r[i]] == c[l[i]] && b[l[i]] == 1 && b[r[i]] == 1) tt = -1;
    mmax = max(mmax, tt);
  }
  cout << ans + mmax << '\n';
  for (int i = 0; i < n; i++) {
    add(l[i], -1);
    add(r[i] + 1, 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> t;
  while (t--) solve();
  return 0;
}

```