#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long double Ma_PI = 3.1415926535897932384626433832795028841971L;
void init() {
  iostream::sync_with_stdio(false);
  cout << fixed << setprecision(20);
}
unsigned long xor128() {
  static unsigned long x = 123456789, y = 362436069, z = 521288629,
                       w = 88675123;
  unsigned long t = (x ^ (x << 11));
  x = y;
  y = z;
  z = w;
  return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
long long imos[2000000];
long long cnt[2000000];
void solve() {
  map<long long, long long> zipper;
  long long n;
  cin >> n;
  vector<pair<long long, long long>> inputs(n);
  for (long long i = 0; i < n; ++i) {
    cin >> inputs[i].first >> inputs[i].second;
    inputs[i].first *= 2LL;
    inputs[i].second *= 2LL;
    zipper[inputs[i].first] = 1;
    zipper[inputs[i].second] = 1;
    zipper[inputs[i].second + 1] = 1;
  }
  long long now_itr = 0;
  for (auto x = zipper.begin(); x != zipper.end(); ++x) {
    x->second = now_itr;
    now_itr++;
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    imos[i] = 0;
    cnt[i] = 0;
  }
  for (long long i = 0; i < n; ++i) {
    imos[zipper[inputs[i].first]]++;
    imos[zipper[inputs[i].second + 1]]--;
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    imos[i + 1] += imos[i];
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    if (imos[i] == 1) {
      if (i == 0 || imos[i - 1] != 1) {
        cnt[i]++;
      }
    }
    cnt[i + 1] += cnt[i];
  }
  long long back = 0;
  vector<long long> right_move, left_move;
  for (long long i = now_itr; i >= 0; --i) {
    if (imos[i] > 1) {
      if (back == 0) {
        right_move.push_back(i);
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  back = 0;
  for (long long i = 0; i < now_itr + 1; ++i) {
    if (imos[i] > 1) {
      if (back == 0) {
        left_move.push_back(i);
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  sort((left_move).begin(), (left_move).end());
  sort((right_move).begin(), (right_move).end());
  right_move.push_back(1e9);
  left_move.push_back(1e9);
  long long base_ans = 0;
  back = 0;
  for (long long i = 0; i < now_itr + 1; ++i) {
    if (imos[i] > 0) {
      if (back == 0) {
        base_ans++;
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  long long ans = -1;
  for (long long i = 0; i < inputs.size(); ++i) {
    long long left = zipper[inputs[i].first];
    long long right = zipper[inputs[i].second];
    long long tmp_ans = 0;
    auto x = lower_bound((right_move).begin(), (right_move).end(), left);
    if (imos[left] == 1) {
      if (left >= 1 && imos[left - 1] >= 1) {
        tmp_ans++;
      }
      if (*x > right) {
        if (imos[right + 1] == 0) {
          tmp_ans--;
        }
        ans = max(ans, tmp_ans);
        continue;
      }
    }
    left = *x;
    x = lower_bound((left_move).begin(), (left_move).end(), right + 1);
    x--;
    if (imos[right] == 1) {
      if (imos[right + 1] >= 1) {
        tmp_ans++;
      }
    }
    right = *x;
    long long bonus = cnt[right] - cnt[left];
    ans = max(ans, tmp_ans + bonus);
  }
  cout << base_ans + ans << endl;
}
int main() {
  init();
  int t;
  cin >> t;
  for (long long te = 0; te < t; ++te) solve();
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long double Ma_PI = 3.1415926535897932384626433832795028841971L;
void init() {
  iostream::sync_with_stdio(false);
  cout << fixed << setprecision(20);
}
unsigned long xor128() {
  static unsigned long x = 123456789, y = 362436069, z = 521288629,
                       w = 88675123;
  unsigned long t = (x ^ (x << 11));
  x = y;
  y = z;
  z = w;
  return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
long long imos[2000000];
long long cnt[2000000];
void solve() {
  map<long long, long long> zipper;
  long long n;
  cin >> n;
  vector<pair<long long, long long>> inputs(n);
  for (long long i = 0; i < n; ++i) {
    cin >> inputs[i].first >> inputs[i].second;
    inputs[i].first *= 2LL;
    inputs[i].second *= 2LL;
    zipper[inputs[i].first] = 1;
    zipper[inputs[i].second] = 1;
    zipper[inputs[i].second + 1] = 1;
  }
  long long now_itr = 0;
  for (auto x = zipper.begin(); x != zipper.end(); ++x) {
    x->second = now_itr;
    now_itr++;
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    imos[i] = 0;
    cnt[i] = 0;
  }
  for (long long i = 0; i < n; ++i) {
    imos[zipper[inputs[i].first]]++;
    imos[zipper[inputs[i].second + 1]]--;
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    imos[i + 1] += imos[i];
  }
  for (long long i = 0; i < now_itr + 10; ++i) {
    if (imos[i] == 1) {
      if (i == 0 || imos[i - 1] != 1) {
        cnt[i]++;
      }
    }
    cnt[i + 1] += cnt[i];
  }
  long long back = 0;
  vector<long long> right_move, left_move;
  for (long long i = now_itr; i >= 0; --i) {
    if (imos[i] > 1) {
      if (back == 0) {
        right_move.push_back(i);
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  back = 0;
  for (long long i = 0; i < now_itr + 1; ++i) {
    if (imos[i] > 1) {
      if (back == 0) {
        left_move.push_back(i);
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  sort((left_move).begin(), (left_move).end());
  sort((right_move).begin(), (right_move).end());
  right_move.push_back(1e9);
  left_move.push_back(1e9);
  long long base_ans = 0;
  back = 0;
  for (long long i = 0; i < now_itr + 1; ++i) {
    if (imos[i] > 0) {
      if (back == 0) {
        base_ans++;
        back = 1;
      }
    } else {
      back = 0;
    }
  }
  long long ans = -1;
  for (long long i = 0; i < inputs.size(); ++i) {
    long long left = zipper[inputs[i].first];
    long long right = zipper[inputs[i].second];
    long long tmp_ans = 0;
    auto x = lower_bound((right_move).begin(), (right_move).end(), left);
    if (imos[left] == 1) {
      if (left >= 1 && imos[left - 1] >= 1) {
        tmp_ans++;
      }
      if (*x > right) {
        if (imos[right + 1] == 0) {
          tmp_ans--;
        }
        ans = max(ans, tmp_ans);
        continue;
      }
    }
    left = *x;
    x = lower_bound((left_move).begin(), (left_move).end(), right + 1);
    x--;
    if (imos[right] == 1) {
      if (imos[right + 1] >= 1) {
        tmp_ans++;
      }
    }
    right = *x;
    long long bonus = cnt[right] - cnt[left];
    ans = max(ans, tmp_ans + bonus);
  }
  cout << base_ans + ans << endl;
}
int main() {
  init();
  int t;
  cin >> t;
  for (long long te = 0; te < t; ++te) solve();
}

```