#include <bits/stdc++.h>
bool debug = false;
using namespace std;
constexpr int kN = int(2E5 + 10);
int l[kN], r[kN], cnt[kN];
bool went[kN];
void solve() {
  int n, op = 0, ip = 0, ans = 0, tmp = 0, ladd = 0;
  bool f = false;
  queue<int> q;
  vector<int> x;
  vector<pair<int, int>> in, out;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &l[i], &r[i]);
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  for (int i = 1; i <= n; i++) went[i] = false;
  for (int i = 1; i <= n; i++) x.push_back(l[i]);
  for (int i = 1; i <= n; i++) x.push_back(r[i]);
  for (int i = 1; i <= n; i++) in.push_back({l[i], i});
  for (int i = 1; i <= n; i++) out.push_back({r[i], i});
  sort(in.begin(), in.end());
  sort(out.begin(), out.end());
  sort(x.begin(), x.end());
  x.resize(unique(x.begin(), x.end()) - x.begin());
  for (int i : x) {
    while (ip < in.size()) {
      if (in[ip].first <= i)
        q.push(in[ip++].second);
      else
        break;
    }
    f = false;
    if (ip - op >= 2) f = true;
    while (op < out.size()) {
      if (out[op].first <= i)
        went[out[op++].second] = true;
      else
        break;
    }
    if (debug)
      printf("i = %d, ip = %d, op = %d, f = %d\n", i, ip, op, f ? 1 : 0);
    if (ladd != 0 && ip - op == 0 && !f) cnt[ladd]--;
    ladd = 0;
    if (ip - op == 0)
      ans++;
    else if (ip - op == 1 && f) {
      while (!q.empty()) {
        if (went[q.front()])
          q.pop();
        else
          break;
      }
      cnt[q.front()]++;
      ladd = q.front();
    }
  }
  for (int i = 1; i <= n; i++) tmp = max(tmp, cnt[i]);
  if (ans + tmp == n) ans--;
  if (debug) printf("ans = %d, tmp = %d\n", ans, tmp);
  printf("%d\n", ans + tmp);
  return;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

### 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>
bool debug = false;
using namespace std;
constexpr int kN = int(2E5 + 10);
int l[kN], r[kN], cnt[kN];
bool went[kN];
void solve() {
  int n, op = 0, ip = 0, ans = 0, tmp = 0, ladd = 0;
  bool f = false;
  queue<int> q;
  vector<int> x;
  vector<pair<int, int>> in, out;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &l[i], &r[i]);
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  for (int i = 1; i <= n; i++) went[i] = false;
  for (int i = 1; i <= n; i++) x.push_back(l[i]);
  for (int i = 1; i <= n; i++) x.push_back(r[i]);
  for (int i = 1; i <= n; i++) in.push_back({l[i], i});
  for (int i = 1; i <= n; i++) out.push_back({r[i], i});
  sort(in.begin(), in.end());
  sort(out.begin(), out.end());
  sort(x.begin(), x.end());
  x.resize(unique(x.begin(), x.end()) - x.begin());
  for (int i : x) {
    while (ip < in.size()) {
      if (in[ip].first <= i)
        q.push(in[ip++].second);
      else
        break;
    }
    f = false;
    if (ip - op >= 2) f = true;
    while (op < out.size()) {
      if (out[op].first <= i)
        went[out[op++].second] = true;
      else
        break;
    }
    if (debug)
      printf("i = %d, ip = %d, op = %d, f = %d\n", i, ip, op, f ? 1 : 0);
    if (ladd != 0 && ip - op == 0 && !f) cnt[ladd]--;
    ladd = 0;
    if (ip - op == 0)
      ans++;
    else if (ip - op == 1 && f) {
      while (!q.empty()) {
        if (went[q.front()])
          q.pop();
        else
          break;
      }
      cnt[q.front()]++;
      ladd = q.front();
    }
  }
  for (int i = 1; i <= n; i++) tmp = max(tmp, cnt[i]);
  if (ans + tmp == n) ans--;
  if (debug) printf("ans = %d, tmp = %d\n", ans, tmp);
  printf("%d\n", ans + tmp);
  return;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

```