#include <bits/stdc++.h>
using namespace std;
const int MAXN = 800005;
int l[MAXN], r[MAXN], p[MAXN];
int a[MAXN], t[MAXN], ps[MAXN], pst[MAXN];
inline int in() {
  int x, y;
  y = scanf("%d", &x);
  return x;
}
void solve() {
  int n = in();
  map<int, int> mp;
  for (int i = 0; i < 4 * n + 4; i++) a[i] = t[i] = ps[i] = pst[i] = 0;
  for (int i = 0; i < n; i++) {
    l[i] = in(), r[i] = in();
    p[2 * i] = l[i], p[2 * i + 1] = r[i];
  }
  int m = 2 * n;
  sort(p, p + m);
  m = unique(p, p + m) - p;
  for (int i = 0; i < m; i++) mp[p[i]] = i;
  for (int i = 0; i < n; i++) {
    a[mp[l[i]] * 2]++;
    a[mp[r[i]] * 2 + 1]--;
  }
  int interval = 0;
  bool flag = false;
  for (int i = 0; i < 2 * m; i++) {
    ps[i] = a[i];
    if (i) ps[i] += ps[i - 1];
    if (ps[i] > 1) flag = true;
    if (i % 2 == 1) {
      if (ps[i - 1] > 1 && ps[i] == 1) t[i - 1]++;
      if (ps[i - 1] >= 1 && ps[i] == 0) interval++;
    }
  }
  if (!flag) {
    printf("%d\n", n - 1);
    return;
  }
  for (int i = 0; i < 2 * m; i++) {
    pst[i] = t[i];
    if (i) pst[i] += pst[i - 1];
  }
  int MAX = 0;
  for (int i = 0; i < n; i++) {
    int x = mp[l[i]] * 2 - 1, y = mp[r[i]] * 2 - 1;
    int tmp = ((y < 0) ? 0 : pst[y]) - ((x < 0) ? 0 : pst[x]);
    if (ps[y + 1] == 1) tmp--;
    MAX = max(MAX, tmp);
  }
  printf("%d\n", interval + MAX);
}
int main() {
  int t = in();
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int MAXN = 800005;
int l[MAXN], r[MAXN], p[MAXN];
int a[MAXN], t[MAXN], ps[MAXN], pst[MAXN];
inline int in() {
  int x, y;
  y = scanf("%d", &x);
  return x;
}
void solve() {
  int n = in();
  map<int, int> mp;
  for (int i = 0; i < 4 * n + 4; i++) a[i] = t[i] = ps[i] = pst[i] = 0;
  for (int i = 0; i < n; i++) {
    l[i] = in(), r[i] = in();
    p[2 * i] = l[i], p[2 * i + 1] = r[i];
  }
  int m = 2 * n;
  sort(p, p + m);
  m = unique(p, p + m) - p;
  for (int i = 0; i < m; i++) mp[p[i]] = i;
  for (int i = 0; i < n; i++) {
    a[mp[l[i]] * 2]++;
    a[mp[r[i]] * 2 + 1]--;
  }
  int interval = 0;
  bool flag = false;
  for (int i = 0; i < 2 * m; i++) {
    ps[i] = a[i];
    if (i) ps[i] += ps[i - 1];
    if (ps[i] > 1) flag = true;
    if (i % 2 == 1) {
      if (ps[i - 1] > 1 && ps[i] == 1) t[i - 1]++;
      if (ps[i - 1] >= 1 && ps[i] == 0) interval++;
    }
  }
  if (!flag) {
    printf("%d\n", n - 1);
    return;
  }
  for (int i = 0; i < 2 * m; i++) {
    pst[i] = t[i];
    if (i) pst[i] += pst[i - 1];
  }
  int MAX = 0;
  for (int i = 0; i < n; i++) {
    int x = mp[l[i]] * 2 - 1, y = mp[r[i]] * 2 - 1;
    int tmp = ((y < 0) ? 0 : pst[y]) - ((x < 0) ? 0 : pst[x]);
    if (ps[y + 1] == 1) tmp--;
    MAX = max(MAX, tmp);
  }
  printf("%d\n", interval + MAX);
}
int main() {
  int t = in();
  while (t--) solve();
  return 0;
}

```