#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x < y ? x : y;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
const int N = 500010, inf = 2e9;
struct Node {
  int l, r;
  bool operator<(const Node &a) const { return l == a.l ? r < a.r : l < a.l; }
} a[N];
struct Info {
  int x, id, v;
  bool operator<(const Info &b) const { return x < b.x; }
} b[N];
void solve() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) read(a[i].l), read(a[i].r);
  sort(a + 1, a + n + 1);
  int tn = 0, tl = -inf, tr = tl, tot = 0;
  unordered_map<int, int> tmp, sum;
  for (int i = 1; i <= n; i++) {
    if (a[i].r <= tr) continue;
    if (a[i].l > tr) tot++, tl = a[i].l, tmp[tl] = 1;
    chkmax(tr, a[i].r);
  }
  tmp[tl] = 1;
  int m = 0, ans = -inf;
  static int f[N];
  for (int i = 1; i <= n; i++)
    b[++m] = (Info){a[i].l, i, 1}, b[++m] = (Info){a[i].r, i, -1}, f[i] = 0;
  sort(b + 1, b + m + 1);
  set<int> s;
  for (int i = 1; i <= m; i++) {
    int j = i - 1;
    vector<int> v, del;
    v.clear(), del.clear();
    while (j < m && b[j + 1].x == b[i].x)
      if (b[++j].v == 1)
        v.push_back(b[j].id);
      else
        del.push_back(b[j].id);
    if (s.size() == 1 && !v.empty()) f[*s.begin()]++;
    for (int x : v) s.insert(x);
    for (int x : del) s.erase(x);
    i = j;
  }
  for (int i = 1; i <= n; i++)
    if (tmp[a[i].l]) sum[a[i].l]++;
  for (int i = 1; i <= n; i++) chkmax(ans, f[i] - (sum[a[i].l] == 1));
  cout << ans + tot << '\n';
}
int main() {
  int T;
  read(T);
  while (T--) solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x < y ? x : y;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
const int N = 500010, inf = 2e9;
struct Node {
  int l, r;
  bool operator<(const Node &a) const { return l == a.l ? r < a.r : l < a.l; }
} a[N];
struct Info {
  int x, id, v;
  bool operator<(const Info &b) const { return x < b.x; }
} b[N];
void solve() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) read(a[i].l), read(a[i].r);
  sort(a + 1, a + n + 1);
  int tn = 0, tl = -inf, tr = tl, tot = 0;
  unordered_map<int, int> tmp, sum;
  for (int i = 1; i <= n; i++) {
    if (a[i].r <= tr) continue;
    if (a[i].l > tr) tot++, tl = a[i].l, tmp[tl] = 1;
    chkmax(tr, a[i].r);
  }
  tmp[tl] = 1;
  int m = 0, ans = -inf;
  static int f[N];
  for (int i = 1; i <= n; i++)
    b[++m] = (Info){a[i].l, i, 1}, b[++m] = (Info){a[i].r, i, -1}, f[i] = 0;
  sort(b + 1, b + m + 1);
  set<int> s;
  for (int i = 1; i <= m; i++) {
    int j = i - 1;
    vector<int> v, del;
    v.clear(), del.clear();
    while (j < m && b[j + 1].x == b[i].x)
      if (b[++j].v == 1)
        v.push_back(b[j].id);
      else
        del.push_back(b[j].id);
    if (s.size() == 1 && !v.empty()) f[*s.begin()]++;
    for (int x : v) s.insert(x);
    for (int x : del) s.erase(x);
    i = j;
  }
  for (int i = 1; i <= n; i++)
    if (tmp[a[i].l]) sum[a[i].l]++;
  for (int i = 1; i <= n; i++) chkmax(ans, f[i] - (sum[a[i].l] == 1));
  cout << ans + tot << '\n';
}
int main() {
  int T;
  read(T);
  while (T--) solve();
  return 0;
}

```