#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
const long double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while ((ch < '0') || (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
namespace My_Math {
int fac[100000 + 100], invfac[100000 + 100];
int add(int x, int y) { return x + y >= 998244353 ? x + y - 998244353 : x + y; }
int dec(int x, int y) { return x < y ? x - y + 998244353 : x - y; }
int mul(int x, int y) { return 1ll * x * y % 998244353; }
long long qpow(long long x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
int inv(int x) { return qpow(x, 998244353 - 2); }
int C(int n, int m) {
  if ((n < m) || (n < 0) || (m < 0)) return 0;
  return mul(mul(fac[n], invfac[m]), invfac[n - m]);
}
int math_init() {
  fac[0] = invfac[0] = 1;
  for (register int i = 1; i <= 100000; i++) fac[i] = mul(fac[i - 1], i);
  invfac[100000] = inv(fac[100000]);
  for (register int i = 100000 - 1; i >= 1; i--)
    invfac[i] = mul(invfac[i + 1], i + 1);
}
}  // namespace My_Math
using namespace My_Math;
int n, cnt[200200];
pair<int, int> p[500500];
multiset<int> s;
int main() {
  int T = read();
  while (T--) {
    n = read();
    int tot = 0;
    for (register int i = 1; i <= n; i++) {
      p[++tot] = make_pair(read(), -i);
      p[++tot] = make_pair(read(), i);
    }
    sort(p + 1, p + 1 + tot);
    int ans = 0;
    for (register int i = 1; i <= tot; i++) {
      if (p[i].second > 0)
        s.erase(p[i].second);
      else
        s.insert(-p[i].second);
      if (s.empty()) ans++;
      if ((s.size() == 1) && (p[i].second > 0) && (p[i + 1].second < 0) &&
          (p[i].first < p[i + 1].first))
        cnt[*s.begin()]++;
      if ((s.size() == 1) && (p[i].second < 0) && (p[i + 1].second > 0))
        cnt[*s.begin()]--;
    }
    int fin = 0;
    for (register int i = 1; i <= n; i++) fin = max(fin, ans + cnt[i]);
    printf("%d\n", fin);
    for (register int i = 1; i <= n; i++) cnt[i] = 0;
  }
  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;
const int N = 100000;
const long double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while ((ch < '0') || (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
namespace My_Math {
int fac[100000 + 100], invfac[100000 + 100];
int add(int x, int y) { return x + y >= 998244353 ? x + y - 998244353 : x + y; }
int dec(int x, int y) { return x < y ? x - y + 998244353 : x - y; }
int mul(int x, int y) { return 1ll * x * y % 998244353; }
long long qpow(long long x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
int inv(int x) { return qpow(x, 998244353 - 2); }
int C(int n, int m) {
  if ((n < m) || (n < 0) || (m < 0)) return 0;
  return mul(mul(fac[n], invfac[m]), invfac[n - m]);
}
int math_init() {
  fac[0] = invfac[0] = 1;
  for (register int i = 1; i <= 100000; i++) fac[i] = mul(fac[i - 1], i);
  invfac[100000] = inv(fac[100000]);
  for (register int i = 100000 - 1; i >= 1; i--)
    invfac[i] = mul(invfac[i + 1], i + 1);
}
}  // namespace My_Math
using namespace My_Math;
int n, cnt[200200];
pair<int, int> p[500500];
multiset<int> s;
int main() {
  int T = read();
  while (T--) {
    n = read();
    int tot = 0;
    for (register int i = 1; i <= n; i++) {
      p[++tot] = make_pair(read(), -i);
      p[++tot] = make_pair(read(), i);
    }
    sort(p + 1, p + 1 + tot);
    int ans = 0;
    for (register int i = 1; i <= tot; i++) {
      if (p[i].second > 0)
        s.erase(p[i].second);
      else
        s.insert(-p[i].second);
      if (s.empty()) ans++;
      if ((s.size() == 1) && (p[i].second > 0) && (p[i + 1].second < 0) &&
          (p[i].first < p[i + 1].first))
        cnt[*s.begin()]++;
      if ((s.size() == 1) && (p[i].second < 0) && (p[i + 1].second > 0))
        cnt[*s.begin()]--;
    }
    int fin = 0;
    for (register int i = 1; i <= n; i++) fin = max(fin, ans + cnt[i]);
    printf("%d\n", fin);
    for (register int i = 1; i <= n; i++) cnt[i] = 0;
  }
  return 0;
}

```