#include <bits/stdc++.h>
using namespace std;
int N, SZ, tot, ans, color[200000], cnt[200000];
set<int> S;
pair<int, int> A[200000];
vector<int> x, B, tree[1 << 19];
vector<pair<int, int>> L, R, E;
int find_(int a) {
  while (a != color[a]) a = color[a];
  return a;
}
void union_(int a, int b) {
  a = find_(a);
  b = find_(b);
  if (a == b) return;
  if (cnt[a] > cnt[b]) swap(a, b);
  color[a] = b;
  cnt[b] += cnt[a];
  B.push_back(a);
  tot--;
}
void add_edge(int s, int e, int v) {
  if (s > e) return;
  for (s += SZ, e += SZ; s <= e; s >>= 1, e >>= 1) {
    if (s & 1) tree[s++].push_back(v);
    if (~e & 1) tree[e--].push_back(v);
  }
}
void solve(int bit = 1, int s = 0, int e = SZ - 1) {
  int bsz = B.size(), m = (s + e) >> 1;
  if (N <= s) return;
  for (auto i : tree[bit]) union_(E[i].first, E[i].second);
  if (s == e) {
    ans = max(ans, tot);
    while (B.size() > bsz) {
      cnt[color[B.back()]] -= cnt[B.back()];
      color[B.back()] = B.back();
      B.pop_back();
      tot++;
    }
    return;
  }
  solve(2 * bit, s, m);
  solve(2 * bit + 1, m + 1, e);
  while (B.size() > bsz) {
    cnt[color[B.back()]] -= cnt[B.back()];
    color[B.back()] = B.back();
    B.pop_back();
    tot++;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ((void)0);
  ((void)0);
  ((void)0);
  int T;
  for (cin >> T; T--;) {
    int i = 0, j = 0;
    cin >> N;
    tot = N - 1;
    ans = 0;
    x.clear();
    L.clear();
    R.clear(), E.clear();
    for (SZ = 1; SZ < N; SZ <<= 1)
      ;
    for (int i = 2 * SZ; --i;) tree[i].clear();
    for (int i = 0; i < N; i++) {
      cin >> A[i].first >> A[i].second;
      x.push_back(A[i].first);
      x.push_back(A[i].second);
      cnt[color[i] = i] = 1;
    }
    sort(x.begin(), x.end());
    x.resize(unique(x.begin(), x.end()) - x.begin());
    for (int i = 0; i < N; i++) {
      A[i].first = lower_bound(x.begin(), x.end(), A[i].first) - x.begin();
      A[i].second = lower_bound(x.begin(), x.end(), A[i].second) - x.begin();
      L.emplace_back(A[i].first, i);
      R.emplace_back(A[i].second, i);
    }
    sort(L.begin(), L.end());
    sort(R.begin(), R.end());
    for (int c = 0; c < x.size(); c++) {
      while (i < N && L[i].first == c) {
        auto [t, idx] = L[i++];
        if (S.size())
          E.emplace_back(min(*S.begin(), idx), max(*S.begin(), idx));
        if (S.size() > 1)
          E.emplace_back(min(*S.rbegin(), idx), max(*S.rbegin(), idx));
        S.insert(idx);
      }
      while (j < N && R[j].first == c) {
        auto [t, idx] = R[j++];
        S.erase(idx);
        if (S.size())
          E.emplace_back(min(*S.begin(), idx), max(*S.begin(), idx));
        if (S.size() > 1)
          E.emplace_back(min(*S.rbegin(), idx), max(*S.rbegin(), idx));
      }
    }
    sort(E.begin(), E.end());
    E.resize(unique(E.begin(), E.end()) - E.begin());
    for (int i = 0; i < E.size(); i++) {
      auto [a, b] = E[i];
      add_edge(0, a - 1, i);
      add_edge(a + 1, b - 1, i);
      add_edge(b + 1, N - 1, i);
    }
    solve();
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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;
int N, SZ, tot, ans, color[200000], cnt[200000];
set<int> S;
pair<int, int> A[200000];
vector<int> x, B, tree[1 << 19];
vector<pair<int, int>> L, R, E;
int find_(int a) {
  while (a != color[a]) a = color[a];
  return a;
}
void union_(int a, int b) {
  a = find_(a);
  b = find_(b);
  if (a == b) return;
  if (cnt[a] > cnt[b]) swap(a, b);
  color[a] = b;
  cnt[b] += cnt[a];
  B.push_back(a);
  tot--;
}
void add_edge(int s, int e, int v) {
  if (s > e) return;
  for (s += SZ, e += SZ; s <= e; s >>= 1, e >>= 1) {
    if (s & 1) tree[s++].push_back(v);
    if (~e & 1) tree[e--].push_back(v);
  }
}
void solve(int bit = 1, int s = 0, int e = SZ - 1) {
  int bsz = B.size(), m = (s + e) >> 1;
  if (N <= s) return;
  for (auto i : tree[bit]) union_(E[i].first, E[i].second);
  if (s == e) {
    ans = max(ans, tot);
    while (B.size() > bsz) {
      cnt[color[B.back()]] -= cnt[B.back()];
      color[B.back()] = B.back();
      B.pop_back();
      tot++;
    }
    return;
  }
  solve(2 * bit, s, m);
  solve(2 * bit + 1, m + 1, e);
  while (B.size() > bsz) {
    cnt[color[B.back()]] -= cnt[B.back()];
    color[B.back()] = B.back();
    B.pop_back();
    tot++;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ((void)0);
  ((void)0);
  ((void)0);
  int T;
  for (cin >> T; T--;) {
    int i = 0, j = 0;
    cin >> N;
    tot = N - 1;
    ans = 0;
    x.clear();
    L.clear();
    R.clear(), E.clear();
    for (SZ = 1; SZ < N; SZ <<= 1)
      ;
    for (int i = 2 * SZ; --i;) tree[i].clear();
    for (int i = 0; i < N; i++) {
      cin >> A[i].first >> A[i].second;
      x.push_back(A[i].first);
      x.push_back(A[i].second);
      cnt[color[i] = i] = 1;
    }
    sort(x.begin(), x.end());
    x.resize(unique(x.begin(), x.end()) - x.begin());
    for (int i = 0; i < N; i++) {
      A[i].first = lower_bound(x.begin(), x.end(), A[i].first) - x.begin();
      A[i].second = lower_bound(x.begin(), x.end(), A[i].second) - x.begin();
      L.emplace_back(A[i].first, i);
      R.emplace_back(A[i].second, i);
    }
    sort(L.begin(), L.end());
    sort(R.begin(), R.end());
    for (int c = 0; c < x.size(); c++) {
      while (i < N && L[i].first == c) {
        auto [t, idx] = L[i++];
        if (S.size())
          E.emplace_back(min(*S.begin(), idx), max(*S.begin(), idx));
        if (S.size() > 1)
          E.emplace_back(min(*S.rbegin(), idx), max(*S.rbegin(), idx));
        S.insert(idx);
      }
      while (j < N && R[j].first == c) {
        auto [t, idx] = R[j++];
        S.erase(idx);
        if (S.size())
          E.emplace_back(min(*S.begin(), idx), max(*S.begin(), idx));
        if (S.size() > 1)
          E.emplace_back(min(*S.rbegin(), idx), max(*S.rbegin(), idx));
      }
    }
    sort(E.begin(), E.end());
    E.resize(unique(E.begin(), E.end()) - E.begin());
    for (int i = 0; i < E.size(); i++) {
      auto [a, b] = E[i];
      add_edge(0, a - 1, i);
      add_edge(a + 1, b - 1, i);
      add_edge(b + 1, N - 1, i);
    }
    solve();
    cout << ans << '\n';
  }
  return 0;
}

```