#include <bits/stdc++.h>
using namespace std;
long long ggcd(long long x, long long y) {
  while (y) {
    x %= y;
    swap(x, y);
  }
  return x;
}
struct Node {
  int add, min, kol01, l, r;
};
vector<Node> tree;
void addOnSegment(int v, int l, int r, int add) {
  tree[v].add += add;
  tree[v].min += add;
  tree[v].l += add;
  tree[v].r += add;
}
void recalc(int v, int l, int r) {
  tree[v].add = 0;
  tree[v].l = tree[v * 2].l;
  tree[v].r = tree[v * 2 + 1].r;
  tree[v].min = min(tree[2 * v].min, tree[2 * v + 1].min);
  tree[v].kol01 = 0;
  if (tree[2 * v].min == tree[v].min) tree[v].kol01 += tree[2 * v].kol01;
  if (tree[2 * v + 1].min == tree[v].min)
    tree[v].kol01 += tree[2 * v + 1].kol01;
  if (tree[2 * v].r == tree[v].min && tree[2 * v + 1].l != tree[v].min)
    ++tree[v].kol01;
}
void push(int v, int l, int r) {
  if (tree[v].add == 0 || l == r) return;
  int mid = (l + r) / 2;
  addOnSegment(2 * v, l, mid, tree[v].add);
  addOnSegment(2 * v + 1, mid + 1, r, tree[v].add);
  tree[v].add = 0;
}
void add(int v, int l, int r, int ql, int qr, int a) {
  if (qr < l || r < ql) return;
  if (ql <= l && r <= qr) {
    addOnSegment(v, l, r, a);
    return;
  }
  push(v, l, r);
  int mid = (l + r) / 2;
  add(2 * v, l, mid, ql, qr, a);
  add(2 * v + 1, mid + 1, r, ql, qr, a);
  recalc(v, l, r);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > ps;
  vector<int> xs;
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    l *= 2;
    r *= 2;
    xs.push_back(l);
    xs.push_back(l - 1);
    xs.push_back(l + 1);
    xs.push_back(r);
    xs.push_back(r - 1);
    xs.push_back(r + 1);
    ps.push_back(make_pair(l, r));
  }
  sort(xs.begin(), xs.end());
  xs.resize(unique(xs.begin(), xs.end()) - xs.begin());
  for (auto& p : ps) {
    p.first = upper_bound(xs.begin(), xs.end(), p.first) - xs.begin();
    p.second = upper_bound(xs.begin(), xs.end(), p.second) - xs.begin();
  }
  int deg = 1;
  while (deg <= xs.size()) deg += deg;
  tree.assign(deg * 2, Node{0, 0, 0, 0, 0});
  for (int i = 0; i < n; ++i) add(1, 0, deg - 1, ps[i].first, ps[i].second, 1);
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    add(1, 0, deg - 1, ps[i].first, ps[i].second, -1);
    ans = max(ans, tree[1].kol01);
    assert(tree[1].min == 0);
    add(1, 0, deg - 1, ps[i].first, ps[i].second, 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int test;
  cin >> test;
  while (test--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
long long ggcd(long long x, long long y) {
  while (y) {
    x %= y;
    swap(x, y);
  }
  return x;
}
struct Node {
  int add, min, kol01, l, r;
};
vector<Node> tree;
void addOnSegment(int v, int l, int r, int add) {
  tree[v].add += add;
  tree[v].min += add;
  tree[v].l += add;
  tree[v].r += add;
}
void recalc(int v, int l, int r) {
  tree[v].add = 0;
  tree[v].l = tree[v * 2].l;
  tree[v].r = tree[v * 2 + 1].r;
  tree[v].min = min(tree[2 * v].min, tree[2 * v + 1].min);
  tree[v].kol01 = 0;
  if (tree[2 * v].min == tree[v].min) tree[v].kol01 += tree[2 * v].kol01;
  if (tree[2 * v + 1].min == tree[v].min)
    tree[v].kol01 += tree[2 * v + 1].kol01;
  if (tree[2 * v].r == tree[v].min && tree[2 * v + 1].l != tree[v].min)
    ++tree[v].kol01;
}
void push(int v, int l, int r) {
  if (tree[v].add == 0 || l == r) return;
  int mid = (l + r) / 2;
  addOnSegment(2 * v, l, mid, tree[v].add);
  addOnSegment(2 * v + 1, mid + 1, r, tree[v].add);
  tree[v].add = 0;
}
void add(int v, int l, int r, int ql, int qr, int a) {
  if (qr < l || r < ql) return;
  if (ql <= l && r <= qr) {
    addOnSegment(v, l, r, a);
    return;
  }
  push(v, l, r);
  int mid = (l + r) / 2;
  add(2 * v, l, mid, ql, qr, a);
  add(2 * v + 1, mid + 1, r, ql, qr, a);
  recalc(v, l, r);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > ps;
  vector<int> xs;
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    l *= 2;
    r *= 2;
    xs.push_back(l);
    xs.push_back(l - 1);
    xs.push_back(l + 1);
    xs.push_back(r);
    xs.push_back(r - 1);
    xs.push_back(r + 1);
    ps.push_back(make_pair(l, r));
  }
  sort(xs.begin(), xs.end());
  xs.resize(unique(xs.begin(), xs.end()) - xs.begin());
  for (auto& p : ps) {
    p.first = upper_bound(xs.begin(), xs.end(), p.first) - xs.begin();
    p.second = upper_bound(xs.begin(), xs.end(), p.second) - xs.begin();
  }
  int deg = 1;
  while (deg <= xs.size()) deg += deg;
  tree.assign(deg * 2, Node{0, 0, 0, 0, 0});
  for (int i = 0; i < n; ++i) add(1, 0, deg - 1, ps[i].first, ps[i].second, 1);
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    add(1, 0, deg - 1, ps[i].first, ps[i].second, -1);
    ans = max(ans, tree[1].kol01);
    assert(tree[1].min == 0);
    add(1, 0, deg - 1, ps[i].first, ps[i].second, 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int test;
  cin >> test;
  while (test--) solve();
  return 0;
}

```