#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1e6 + 10;
std::map<int, int> m1, m2;
int n, cnt, l[MaxN], r[MaxN], a[MaxN], s[MaxN];
inline void prework() {
  m1.clear(), m2.clear();
  for (int i = 1; i <= n; i++) m1[l[i]] = m1[r[i]] = 1;
  cnt = 0;
  for (std::map<int, int>::iterator it = m1.begin(); it != m1.end(); it++)
    m2[it->first] = ++cnt;
  for (int i = 1; i <= n; i++) {
    l[i] = m2[l[i]] * 2 - 1, r[i] = m2[r[i]] * 2 - 1;
  }
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? x : (-x);
}
int main() {
  int T = read();
  while (T--) {
    n = read(), cnt = 0;
    for (int i = 1; i <= n; i++) l[i] = read(), r[i] = read();
    prework(), cnt = cnt * 2 - 1;
    for (int i = 1; i <= n; i++) a[l[i]]++, a[r[i] + 1]--;
    for (int i = 1; i <= cnt; i++) a[i] += a[i - 1];
    int num = 0, ans = -1000000;
    for (int i = 1; i <= cnt; i++) {
      s[i] = s[i - 1];
      num += (a[i] && !a[i - 1]);
      if (a[i] > 1 && a[i - 1] <= 1) ++s[i];
    }
    for (int i = 1; i <= n; i++) {
      int cur =
          s[r[i]] - s[l[i] - 1] + ((a[l[i]] > 1) && (a[l[i] - 1] > 1)) - 1;
      ans = std::max(ans, cur);
    }
    printf("%d\n", num + ans);
    for (int i = 0; i <= cnt * 2; i++) a[i] = s[i] = 0;
  }
  return 0;
}

### Prompt

Generate 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 = 1e6 + 10;
std::map<int, int> m1, m2;
int n, cnt, l[MaxN], r[MaxN], a[MaxN], s[MaxN];
inline void prework() {
  m1.clear(), m2.clear();
  for (int i = 1; i <= n; i++) m1[l[i]] = m1[r[i]] = 1;
  cnt = 0;
  for (std::map<int, int>::iterator it = m1.begin(); it != m1.end(); it++)
    m2[it->first] = ++cnt;
  for (int i = 1; i <= n; i++) {
    l[i] = m2[l[i]] * 2 - 1, r[i] = m2[r[i]] * 2 - 1;
  }
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? x : (-x);
}
int main() {
  int T = read();
  while (T--) {
    n = read(), cnt = 0;
    for (int i = 1; i <= n; i++) l[i] = read(), r[i] = read();
    prework(), cnt = cnt * 2 - 1;
    for (int i = 1; i <= n; i++) a[l[i]]++, a[r[i] + 1]--;
    for (int i = 1; i <= cnt; i++) a[i] += a[i - 1];
    int num = 0, ans = -1000000;
    for (int i = 1; i <= cnt; i++) {
      s[i] = s[i - 1];
      num += (a[i] && !a[i - 1]);
      if (a[i] > 1 && a[i - 1] <= 1) ++s[i];
    }
    for (int i = 1; i <= n; i++) {
      int cur =
          s[r[i]] - s[l[i] - 1] + ((a[l[i]] > 1) && (a[l[i] - 1] > 1)) - 1;
      ans = std::max(ans, cur);
    }
    printf("%d\n", num + ans);
    for (int i = 0; i <= cnt * 2; i++) a[i] = s[i] = 0;
  }
  return 0;
}

```