#include <bits/stdc++.h>
using namespace std;
const int N = 202020, M = N * 2;
int T, n;
vector<vector<int> > seg;
int cnt[N], rmost[N];
int bit[M];
vector<int> dic;
void upd(int x, int y) {
  for (; x < M; x += x & -x) bit[x] += y;
}
int qry(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret += bit[x];
  return ret;
}
int qry(int l, int r) { return qry(r) - qry(l); }
int gid(int x) { return lower_bound(dic.begin(), dic.end(), x) - dic.begin(); }
int solve() {
  dic.push_back(INT_MIN);
  sort(dic.begin(), dic.end());
  dic.erase(unique(dic.begin(), dic.end()), dic.end());
  sort(seg.begin(), seg.end());
  int mr = INT_MIN;
  for (int i = (0); i < ((int)seg.size()); i++) {
    int l = seg[i][0], r = seg[i][1];
    cnt[i + 1] = cnt[i];
    if (l > mr) {
      cnt[i + 1]++;
    }
    mr = max(mr, r);
    rmost[i + 1] = mr;
  }
  multiset<int> q;
  int ans = 1;
  for (int i = (int)seg.size() - 1; ~i; i--) {
    int p = i - 1;
    if (p == -1) {
      ans = max(ans, qry(M - 1));
    } else {
      int r = gid(rmost[p + 1]);
      ans = max(ans, cnt[p + 1] + qry(r, M - 1));
    }
    int l = gid(seg[i][0]), r = gid(seg[i][1]);
    while ((int)q.size() && *q.begin() <= r) {
      upd(*q.begin(), -1);
      q.erase(q.begin());
    }
    upd(l, 1);
    q.insert(l);
  }
  while ((int)q.size()) {
    upd(*q.begin(), -1);
    q.erase(q.begin());
  }
  return ans;
}
int main() {
  scanf("%d", &T);
  for (int cas = (0); cas < (T); cas++) {
    scanf("%d", &n);
    seg.clear();
    dic.clear();
    for (int i = (0); i < (n); i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      dic.push_back(l);
      dic.push_back(r);
      seg.push_back({l, r});
    }
    printf("%d\n", solve());
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 = 202020, M = N * 2;
int T, n;
vector<vector<int> > seg;
int cnt[N], rmost[N];
int bit[M];
vector<int> dic;
void upd(int x, int y) {
  for (; x < M; x += x & -x) bit[x] += y;
}
int qry(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret += bit[x];
  return ret;
}
int qry(int l, int r) { return qry(r) - qry(l); }
int gid(int x) { return lower_bound(dic.begin(), dic.end(), x) - dic.begin(); }
int solve() {
  dic.push_back(INT_MIN);
  sort(dic.begin(), dic.end());
  dic.erase(unique(dic.begin(), dic.end()), dic.end());
  sort(seg.begin(), seg.end());
  int mr = INT_MIN;
  for (int i = (0); i < ((int)seg.size()); i++) {
    int l = seg[i][0], r = seg[i][1];
    cnt[i + 1] = cnt[i];
    if (l > mr) {
      cnt[i + 1]++;
    }
    mr = max(mr, r);
    rmost[i + 1] = mr;
  }
  multiset<int> q;
  int ans = 1;
  for (int i = (int)seg.size() - 1; ~i; i--) {
    int p = i - 1;
    if (p == -1) {
      ans = max(ans, qry(M - 1));
    } else {
      int r = gid(rmost[p + 1]);
      ans = max(ans, cnt[p + 1] + qry(r, M - 1));
    }
    int l = gid(seg[i][0]), r = gid(seg[i][1]);
    while ((int)q.size() && *q.begin() <= r) {
      upd(*q.begin(), -1);
      q.erase(q.begin());
    }
    upd(l, 1);
    q.insert(l);
  }
  while ((int)q.size()) {
    upd(*q.begin(), -1);
    q.erase(q.begin());
  }
  return ans;
}
int main() {
  scanf("%d", &T);
  for (int cas = (0); cas < (T); cas++) {
    scanf("%d", &n);
    seg.clear();
    dic.clear();
    for (int i = (0); i < (n); i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      dic.push_back(l);
      dic.push_back(r);
      seg.push_back({l, r});
    }
    printf("%d\n", solve());
  }
  return 0;
}

```