#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int t, n, l[maxn], r[maxn], cnt, re[maxn], m, ans;
bool vis[maxn], qs[maxn];
vector<int> ps, s[2 * maxn], e[2 * maxn];
set<int> st;
int main() {
  cin >> t;
  while (t--) {
    ps.clear();
    st.clear();
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> l[i] >> r[i];
      r[i]++;
      ps.push_back(l[i]);
      ps.push_back(r[i]);
    }
    sort(ps.begin(), ps.end());
    ps.resize(unique(ps.begin(), ps.end()) - ps.begin());
    for (int i = 0; i < ps.size(); i++) {
      s[i].clear();
      e[i].clear();
    }
    for (int i = 0; i < n; i++) {
      l[i] = lower_bound(ps.begin(), ps.end(), l[i]) - ps.begin();
      r[i] = lower_bound(ps.begin(), ps.end(), r[i]) - ps.begin();
      s[l[i]].push_back(i);
      e[r[i]].push_back(i);
      vis[i] = 0;
      re[i] = 0;
      qs[i] = 0;
    }
    cnt = 0;
    m = 0;
    for (int i = 0; i < ps.size(); i++) {
      int cntl = cnt, il = -1, cntp = 2, il2 = -1;
      if (cnt == 1) il = (*st.begin());
      for (int u : e[i]) st.erase(u);
      cntp = st.size();
      if (cntp == 1) il2 = (*st.begin());
      for (int u : s[i]) {
        if (st.size() == 1) {
          qs[(*st.begin())] = 1;
        }
        st.insert(u);
        if (st.size() > 1) qs[u] = 1;
      }
      cnt = st.size();
      if (cnt && cntp == 0) m++;
      if (cntl > 1 && cnt == 1) {
        int x = (*st.begin());
        if (l[x] != i) {
          vis[x] = 1;
        }
      }
      if (cntl == 1 && cnt > 1) {
        if (vis[il] && r[il] != i) {
          re[il]++;
        }
      }
      if (cntl > 1 && cntp == 1 && cnt > 1) {
        vis[il2] = 1;
        re[il2]++;
      }
    }
    ans = -1;
    for (int i = 0; i < n; i++) {
      if (qs[i])
        ans = max(ans, re[i] + m);
      else
        ans = max(ans, m - 1);
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const int maxn = 2e5 + 5;
int t, n, l[maxn], r[maxn], cnt, re[maxn], m, ans;
bool vis[maxn], qs[maxn];
vector<int> ps, s[2 * maxn], e[2 * maxn];
set<int> st;
int main() {
  cin >> t;
  while (t--) {
    ps.clear();
    st.clear();
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> l[i] >> r[i];
      r[i]++;
      ps.push_back(l[i]);
      ps.push_back(r[i]);
    }
    sort(ps.begin(), ps.end());
    ps.resize(unique(ps.begin(), ps.end()) - ps.begin());
    for (int i = 0; i < ps.size(); i++) {
      s[i].clear();
      e[i].clear();
    }
    for (int i = 0; i < n; i++) {
      l[i] = lower_bound(ps.begin(), ps.end(), l[i]) - ps.begin();
      r[i] = lower_bound(ps.begin(), ps.end(), r[i]) - ps.begin();
      s[l[i]].push_back(i);
      e[r[i]].push_back(i);
      vis[i] = 0;
      re[i] = 0;
      qs[i] = 0;
    }
    cnt = 0;
    m = 0;
    for (int i = 0; i < ps.size(); i++) {
      int cntl = cnt, il = -1, cntp = 2, il2 = -1;
      if (cnt == 1) il = (*st.begin());
      for (int u : e[i]) st.erase(u);
      cntp = st.size();
      if (cntp == 1) il2 = (*st.begin());
      for (int u : s[i]) {
        if (st.size() == 1) {
          qs[(*st.begin())] = 1;
        }
        st.insert(u);
        if (st.size() > 1) qs[u] = 1;
      }
      cnt = st.size();
      if (cnt && cntp == 0) m++;
      if (cntl > 1 && cnt == 1) {
        int x = (*st.begin());
        if (l[x] != i) {
          vis[x] = 1;
        }
      }
      if (cntl == 1 && cnt > 1) {
        if (vis[il] && r[il] != i) {
          re[il]++;
        }
      }
      if (cntl > 1 && cntp == 1 && cnt > 1) {
        vis[il2] = 1;
        re[il2]++;
      }
    }
    ans = -1;
    for (int i = 0; i < n; i++) {
      if (qs[i])
        ans = max(ans, re[i] + m);
      else
        ans = max(ans, m - 1);
    }
    cout << ans << endl;
  }
  return 0;
}

```