#include <bits/stdc++.h>
using namespace std;
long long v[200005], p[200005], k;
long long fun(long long k, long long x) {
  if (k == 0) return 0;
  long long mid, low, high;
  low = 0;
  high = k - 1;
  long long ans = k;
  while (low <= high) {
    mid = (low + high) / 2;
    if (v[mid] <= x) {
      ans = min(ans, mid);
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, i, j, t, x, y, w, dd = 1;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<pair<long long, long long> > seg;
    vector<long long> m(n), pi(n);
    for (i = 0; i < n; i++) {
      cin >> x >> y;
      seg.push_back({x, y});
    }
    sort(seg.begin(), seg.end());
    long long maxi = -999999999999999;
    for (i = 0; i < n; i++) {
      if (seg[i].first > maxi) {
        m[i] = (i == 0) ? 1 : (m[i - 1] + 1);
      } else {
        m[i] = m[i - 1];
      }
      maxi = max(maxi, seg[i].second);
      if (i != 0)
        p[i] = max(p[i - 1], seg[i].second);
      else
        p[i] = seg[i].second;
    }
    long long ans = 0;
    k = 0;
    for (i = n - 1; i > 0; i--) {
      long long l = seg[i].first;
      long long r = seg[i].second;
      long long temp = fun(k, p[i - 1]);
      ans = max(ans, m[i - 1] + temp);
      if (k > 0) {
        for (j = k - 1; j >= 0; j--) {
          if (v[j] <= r) {
            k--;
          } else {
            break;
          }
        }
      }
      v[k++] = l;
    }
    ans = max(ans, k);
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long v[200005], p[200005], k;
long long fun(long long k, long long x) {
  if (k == 0) return 0;
  long long mid, low, high;
  low = 0;
  high = k - 1;
  long long ans = k;
  while (low <= high) {
    mid = (low + high) / 2;
    if (v[mid] <= x) {
      ans = min(ans, mid);
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, i, j, t, x, y, w, dd = 1;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<pair<long long, long long> > seg;
    vector<long long> m(n), pi(n);
    for (i = 0; i < n; i++) {
      cin >> x >> y;
      seg.push_back({x, y});
    }
    sort(seg.begin(), seg.end());
    long long maxi = -999999999999999;
    for (i = 0; i < n; i++) {
      if (seg[i].first > maxi) {
        m[i] = (i == 0) ? 1 : (m[i - 1] + 1);
      } else {
        m[i] = m[i - 1];
      }
      maxi = max(maxi, seg[i].second);
      if (i != 0)
        p[i] = max(p[i - 1], seg[i].second);
      else
        p[i] = seg[i].second;
    }
    long long ans = 0;
    k = 0;
    for (i = n - 1; i > 0; i--) {
      long long l = seg[i].first;
      long long r = seg[i].second;
      long long temp = fun(k, p[i - 1]);
      ans = max(ans, m[i - 1] + temp);
      if (k > 0) {
        for (j = k - 1; j >= 0; j--) {
          if (v[j] <= r) {
            k--;
          } else {
            break;
          }
        }
      }
      v[k++] = l;
    }
    ans = max(ans, k);
    cout << ans << "\n";
  }
  return 0;
}

```