#include <bits/stdc++.h>
using namespace std;
const int inf = 20;
const int maxx = 1e7;
struct segtree {
  struct node {
    int t, v;
    bool L, R;
    int ls, rs;
    node() {
      v = t = 0;
      ls = rs = 0;
      L = R = 0;
    }
  } t[maxx];
  int tot;
  void update(int u, int k, int l, int r) {
    t[u].t += k;
    if (t[u].t) {
      t[u].v = t[u].L = t[u].R = 1;
    } else {
      if (l == r) {
        t[u].v = t[u].L = t[u].R = 0;
        return;
      }
      pushup(u);
    }
  }
  void pushup(int u) {
    if (t[u].t) return;
    t[u].L = t[t[u].ls].L;
    t[u].R = t[t[u].rs].R;
    t[u].v = t[t[u].ls].v + t[t[u].rs].v - (t[t[u].ls].R && t[t[u].rs].L);
  }
  int range(int &u, int L, int R, int l, int r, int k = 0) {
    if (!u) u = ++tot;
    if (L <= l && r <= R) {
      update(u, k, l, r);
      return t[u].v;
    }
    int ans = 0;
    int m = (l + r) >> 1;
    if (L <= m) ans += range(t[u].ls, L, R, l, m, k);
    if (R > m) ans += range(t[u].rs, L, R, m + 1, r, k);
    pushup(u);
    return ans;
  }
} seg;
int rt = 0;
void solve() {
  int n;
  cin >> n;
  int a[n], b[n];
  vector<int> v;
  for (int i = (0); i < (n); ++i) {
    cin >> a[i] >> b[i];
    a[i] = a[i] * 2;
    b[i] = b[i] * 2;
    v.push_back(a[i]), v.push_back(b[i]), v.push_back(a[i] - 1),
        v.push_back(b[i] + 1);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = (0); i < (n); ++i) {
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
    b[i] = lower_bound(v.begin(), v.end(), b[i]) - v.begin();
  }
  for (int i = (0); i < (n); ++i) seg.range(rt, a[i], b[i], 0, v.size() - 1, 1);
  int ret = 0;
  for (int i = (0); i < (n); ++i) {
    seg.range(rt, a[i], b[i], 0, v.size() - 1, -1);
    ret = max(ret, seg.range(rt, 0, v.size() - 1, 0, v.size() - 1));
    seg.range(rt, a[i], b[i], 0, v.size() - 1, 1);
  }
  for (int i = (1); i < (seg.tot + 1); ++i) seg.t[i] = segtree::node();
  seg.tot = rt = 0;
  cout << ret << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please provide a Cpp coded solution 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 inf = 20;
const int maxx = 1e7;
struct segtree {
  struct node {
    int t, v;
    bool L, R;
    int ls, rs;
    node() {
      v = t = 0;
      ls = rs = 0;
      L = R = 0;
    }
  } t[maxx];
  int tot;
  void update(int u, int k, int l, int r) {
    t[u].t += k;
    if (t[u].t) {
      t[u].v = t[u].L = t[u].R = 1;
    } else {
      if (l == r) {
        t[u].v = t[u].L = t[u].R = 0;
        return;
      }
      pushup(u);
    }
  }
  void pushup(int u) {
    if (t[u].t) return;
    t[u].L = t[t[u].ls].L;
    t[u].R = t[t[u].rs].R;
    t[u].v = t[t[u].ls].v + t[t[u].rs].v - (t[t[u].ls].R && t[t[u].rs].L);
  }
  int range(int &u, int L, int R, int l, int r, int k = 0) {
    if (!u) u = ++tot;
    if (L <= l && r <= R) {
      update(u, k, l, r);
      return t[u].v;
    }
    int ans = 0;
    int m = (l + r) >> 1;
    if (L <= m) ans += range(t[u].ls, L, R, l, m, k);
    if (R > m) ans += range(t[u].rs, L, R, m + 1, r, k);
    pushup(u);
    return ans;
  }
} seg;
int rt = 0;
void solve() {
  int n;
  cin >> n;
  int a[n], b[n];
  vector<int> v;
  for (int i = (0); i < (n); ++i) {
    cin >> a[i] >> b[i];
    a[i] = a[i] * 2;
    b[i] = b[i] * 2;
    v.push_back(a[i]), v.push_back(b[i]), v.push_back(a[i] - 1),
        v.push_back(b[i] + 1);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = (0); i < (n); ++i) {
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
    b[i] = lower_bound(v.begin(), v.end(), b[i]) - v.begin();
  }
  for (int i = (0); i < (n); ++i) seg.range(rt, a[i], b[i], 0, v.size() - 1, 1);
  int ret = 0;
  for (int i = (0); i < (n); ++i) {
    seg.range(rt, a[i], b[i], 0, v.size() - 1, -1);
    ret = max(ret, seg.range(rt, 0, v.size() - 1, 0, v.size() - 1));
    seg.range(rt, a[i], b[i], 0, v.size() - 1, 1);
  }
  for (int i = (1); i < (seg.tot + 1); ++i) seg.t[i] = segtree::node();
  seg.tot = rt = 0;
  cout << ret << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

```