#include <bits/stdc++.h>
using namespace std;
struct emp {
  int val, num;
  bool le, ri;
} tree[4 * 800005], O;
long long n, lef[800005], rig[800005], lazy[4 * 800005], q;
vector<long long> v;
void propagate(int s, int e, int x) {
  if (s != e) {
    lazy[2 * x] += lazy[x];
    lazy[2 * x + 1] += lazy[x];
  }
  tree[x].num += (e - s + 1) * lazy[x];
  lazy[x] = 0;
  return;
}
void update(int s, int e, int x, int l, int r) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (r < s || e < l) return;
  if (l <= s && e <= r) {
    lazy[x]++;
    propagate(s, e, x);
    return;
  }
  update(s, m, 2 * x, l, r);
  update(m + 1, e, 2 * x + 1, l, r);
  tree[x].num = tree[2 * x].num + tree[2 * x + 1].num;
  return;
}
void init(int s, int e, int x) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (s == e) {
    tree[x].le = tree[x].ri = tree[x].val = (tree[x].num <= 1);
    return;
  }
  init(s, m, 2 * x);
  init(m + 1, e, 2 * x + 1);
  tree[x].le = tree[2 * x].le;
  tree[x].ri = tree[2 * x + 1].ri;
  tree[x].val = tree[2 * x].val + tree[2 * x + 1].val -
                (tree[2 * x].ri && tree[2 * x + 1].le);
  tree[x].num = tree[2 * x].num + tree[2 * x + 1].num;
  return;
}
emp read(int s, int e, int x, int l, int r) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (r < s || e < l) return O;
  if (l <= s && e <= r) return tree[x];
  emp p, p1 = read(s, m, 2 * x, l, r), p2 = read(m + 1, e, 2 * x + 1, l, r);
  p.val = p1.val + p2.val - (p1.ri && p2.le);
  p.le = p1.le;
  p.ri = p2.ri;
  p.num = p1.num + p2.num;
  return p;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> lef[i] >> rig[i];
      v.push_back(lef[i]);
      v.push_back(rig[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    int sz = 2 * v.size() - 1;
    for (int i = 0; i < n; i++) {
      lef[i] = 2 * (lower_bound(v.begin(), v.end(), lef[i]) - v.begin());
      rig[i] = 2 * (lower_bound(v.begin(), v.end(), rig[i]) - v.begin());
      update(0, sz - 1, 1, lef[i], rig[i]);
    }
    init(0, sz - 1, 1);
    int maxn = -1e9;
    for (int i = 0; i < n; i++) {
      maxn = max(maxn, read(0, sz - 1, 1, lef[i], rig[i]).val -
                           read(0, sz - 1, 1, lef[i], lef[i]).val -
                           read(0, sz - 1, 1, rig[i], rig[i]).val);
    }
    update(0, sz - 1, 1, 0, sz - 1);
    init(0, sz - 1, 1);
    cout << maxn + read(0, sz - 1, 1, 0, sz - 1).val + 1 << "\n";
    for (int i = 0; i < 4 * sz; i++) {
      tree[i] = O;
      lazy[i] = 0;
    }
    v.clear();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
struct emp {
  int val, num;
  bool le, ri;
} tree[4 * 800005], O;
long long n, lef[800005], rig[800005], lazy[4 * 800005], q;
vector<long long> v;
void propagate(int s, int e, int x) {
  if (s != e) {
    lazy[2 * x] += lazy[x];
    lazy[2 * x + 1] += lazy[x];
  }
  tree[x].num += (e - s + 1) * lazy[x];
  lazy[x] = 0;
  return;
}
void update(int s, int e, int x, int l, int r) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (r < s || e < l) return;
  if (l <= s && e <= r) {
    lazy[x]++;
    propagate(s, e, x);
    return;
  }
  update(s, m, 2 * x, l, r);
  update(m + 1, e, 2 * x + 1, l, r);
  tree[x].num = tree[2 * x].num + tree[2 * x + 1].num;
  return;
}
void init(int s, int e, int x) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (s == e) {
    tree[x].le = tree[x].ri = tree[x].val = (tree[x].num <= 1);
    return;
  }
  init(s, m, 2 * x);
  init(m + 1, e, 2 * x + 1);
  tree[x].le = tree[2 * x].le;
  tree[x].ri = tree[2 * x + 1].ri;
  tree[x].val = tree[2 * x].val + tree[2 * x + 1].val -
                (tree[2 * x].ri && tree[2 * x + 1].le);
  tree[x].num = tree[2 * x].num + tree[2 * x + 1].num;
  return;
}
emp read(int s, int e, int x, int l, int r) {
  int m = s + e >> 1;
  propagate(s, e, x);
  if (r < s || e < l) return O;
  if (l <= s && e <= r) return tree[x];
  emp p, p1 = read(s, m, 2 * x, l, r), p2 = read(m + 1, e, 2 * x + 1, l, r);
  p.val = p1.val + p2.val - (p1.ri && p2.le);
  p.le = p1.le;
  p.ri = p2.ri;
  p.num = p1.num + p2.num;
  return p;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> lef[i] >> rig[i];
      v.push_back(lef[i]);
      v.push_back(rig[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    int sz = 2 * v.size() - 1;
    for (int i = 0; i < n; i++) {
      lef[i] = 2 * (lower_bound(v.begin(), v.end(), lef[i]) - v.begin());
      rig[i] = 2 * (lower_bound(v.begin(), v.end(), rig[i]) - v.begin());
      update(0, sz - 1, 1, lef[i], rig[i]);
    }
    init(0, sz - 1, 1);
    int maxn = -1e9;
    for (int i = 0; i < n; i++) {
      maxn = max(maxn, read(0, sz - 1, 1, lef[i], rig[i]).val -
                           read(0, sz - 1, 1, lef[i], lef[i]).val -
                           read(0, sz - 1, 1, rig[i], rig[i]).val);
    }
    update(0, sz - 1, 1, 0, sz - 1);
    init(0, sz - 1, 1);
    cout << maxn + read(0, sz - 1, 1, 0, sz - 1).val + 1 << "\n";
    for (int i = 0; i < 4 * sz; i++) {
      tree[i] = O;
      lazy[i] = 0;
    }
    v.clear();
  }
  return 0;
}

```