#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
const int maxn = 2e5 + 5;
struct H {
  int x;
  int flag;
} s[maxn << 1];
bool cmp(const H& a, const H& b) {
  if (a.x < b.x) return true;
  if (a.x == b.x) return a.flag > b.flag;
  return false;
}
set<int> st;
int only[maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    st.clear();
    memset((only), (0), sizeof(only));
    int n, cnt = 0;
    cin >> n;
    for (int i = (1); i <= (n); i++) {
      int x, y;
      cin >> x >> y;
      s[++cnt].x = x;
      s[cnt].flag = i;
      s[++cnt].x = y;
      s[cnt].flag = -i;
    }
    sort(s + 1, s + cnt + 1, cmp);
    int ans = 0;
    for (int i = (1); i <= (cnt); i++) {
      if (s[i].flag > 0) {
        if (st.empty()) only[abs(s[i].flag)]--;
        st.insert(abs(s[i].flag));
      } else {
        st.erase(abs(s[i].flag));
        if (st.empty()) only[abs(s[i].flag)]--, ans++;
      }
      if (st.size() == 1) only[*st.begin()]++;
    }
    int MAX = -1;
    for (int i = (1); i <= (n); i++) MAX = max(only[i], MAX);
    cout << ans + MAX << "\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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
const int maxn = 2e5 + 5;
struct H {
  int x;
  int flag;
} s[maxn << 1];
bool cmp(const H& a, const H& b) {
  if (a.x < b.x) return true;
  if (a.x == b.x) return a.flag > b.flag;
  return false;
}
set<int> st;
int only[maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    st.clear();
    memset((only), (0), sizeof(only));
    int n, cnt = 0;
    cin >> n;
    for (int i = (1); i <= (n); i++) {
      int x, y;
      cin >> x >> y;
      s[++cnt].x = x;
      s[cnt].flag = i;
      s[++cnt].x = y;
      s[cnt].flag = -i;
    }
    sort(s + 1, s + cnt + 1, cmp);
    int ans = 0;
    for (int i = (1); i <= (cnt); i++) {
      if (s[i].flag > 0) {
        if (st.empty()) only[abs(s[i].flag)]--;
        st.insert(abs(s[i].flag));
      } else {
        st.erase(abs(s[i].flag));
        if (st.empty()) only[abs(s[i].flag)]--, ans++;
      }
      if (st.size() == 1) only[*st.begin()]++;
    }
    int MAX = -1;
    for (int i = (1); i <= (n); i++) MAX = max(only[i], MAX);
    cout << ans + MAX << "\n";
  }
  return 0;
}

```