#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, ST[4 * N], a[N], s[N];
struct T {
  int l, r;
};
T q[N];
void build(int id, int l, int r) {
  if (l == r) {
    ST[id] = 0;
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  ST[id] = ST[id * 2] + ST[id * 2 + 1];
}
int get(int id, int l, int r, int L, int R) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) {
    return ST[id];
  }
  int mid = (l + r) / 2;
  return get(id * 2, l, mid, L, R) + get(id * 2 + 1, mid + 1, r, L, R);
}
void update(int id, int l, int r, int u, int x) {
  if (r == l && l == u) {
    ST[id] += x;
    return;
  }
  if (u > r || u < l) return;
  if (u <= r && u >= l) {
    int mid = (l + r) / 2;
    update(id * 2, l, mid, u, x);
    update(id * 2 + 1, mid + 1, r, u, x);
    ST[id] = ST[id * 2] + ST[id * 2 + 1];
  }
}
void sol() {
  vector<int> val;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].l *= 2;
    q[i].r *= 2;
    val.push_back(q[i].l);
    val.push_back(q[i].r);
    val.push_back(q[i].l - 1);
    val.push_back(q[i].r + 1);
  }
  sort(val.begin(), val.end());
  val.erase(unique(val.begin(), val.end()), val.end());
  int sz = val.size();
  build(1, 1, sz);
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(val.begin(), val.end(), q[i].l) - val.begin() + 1;
    int r = lower_bound(val.begin(), val.end(), q[i].r) - val.begin() + 1;
    q[i].l = l;
    q[i].r = r;
    update(1, 1, sz, l, 1);
    if (r < sz) update(1, 1, sz, r + 1, -1);
  }
  for (int i = 1; i <= sz; i++) a[i] = get(1, 1, sz, 1, i);
  vector<int> range;
  int i = 1, dem = 0;
  while (i <= sz) {
    range.push_back(i);
    if (a[i] == 0 && i != 1 && i != sz) dem++;
    int j = i;
    while (j <= sz && a[j] == a[i]) j++;
    i = j;
  }
  s[0] = 0;
  for (int i = 1; i < range.size() - 1; i++) {
    s[i] = s[i - 1];
    if (a[range[i]] == 1 && a[range[i - 1]] > 0 && a[range[i + 1]] > 0)
      s[i]++;
    else if (a[range[i]] == 1 && a[range[i - 1]] == 0 && a[range[i + 1]] == 0)
      s[i]--;
  }
  int res = -1e9;
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(range.begin(), range.end(), q[i].l) - range.begin();
    int r = upper_bound(range.begin(), range.end(), q[i].r) - range.begin() - 1;
    int ans = s[r] - s[max(l - 1, 0)];
    res = max(res, ans);
  }
  cout << res + dem + 1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) sol();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e6 + 5;
int n, ST[4 * N], a[N], s[N];
struct T {
  int l, r;
};
T q[N];
void build(int id, int l, int r) {
  if (l == r) {
    ST[id] = 0;
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  ST[id] = ST[id * 2] + ST[id * 2 + 1];
}
int get(int id, int l, int r, int L, int R) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) {
    return ST[id];
  }
  int mid = (l + r) / 2;
  return get(id * 2, l, mid, L, R) + get(id * 2 + 1, mid + 1, r, L, R);
}
void update(int id, int l, int r, int u, int x) {
  if (r == l && l == u) {
    ST[id] += x;
    return;
  }
  if (u > r || u < l) return;
  if (u <= r && u >= l) {
    int mid = (l + r) / 2;
    update(id * 2, l, mid, u, x);
    update(id * 2 + 1, mid + 1, r, u, x);
    ST[id] = ST[id * 2] + ST[id * 2 + 1];
  }
}
void sol() {
  vector<int> val;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].l *= 2;
    q[i].r *= 2;
    val.push_back(q[i].l);
    val.push_back(q[i].r);
    val.push_back(q[i].l - 1);
    val.push_back(q[i].r + 1);
  }
  sort(val.begin(), val.end());
  val.erase(unique(val.begin(), val.end()), val.end());
  int sz = val.size();
  build(1, 1, sz);
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(val.begin(), val.end(), q[i].l) - val.begin() + 1;
    int r = lower_bound(val.begin(), val.end(), q[i].r) - val.begin() + 1;
    q[i].l = l;
    q[i].r = r;
    update(1, 1, sz, l, 1);
    if (r < sz) update(1, 1, sz, r + 1, -1);
  }
  for (int i = 1; i <= sz; i++) a[i] = get(1, 1, sz, 1, i);
  vector<int> range;
  int i = 1, dem = 0;
  while (i <= sz) {
    range.push_back(i);
    if (a[i] == 0 && i != 1 && i != sz) dem++;
    int j = i;
    while (j <= sz && a[j] == a[i]) j++;
    i = j;
  }
  s[0] = 0;
  for (int i = 1; i < range.size() - 1; i++) {
    s[i] = s[i - 1];
    if (a[range[i]] == 1 && a[range[i - 1]] > 0 && a[range[i + 1]] > 0)
      s[i]++;
    else if (a[range[i]] == 1 && a[range[i - 1]] == 0 && a[range[i + 1]] == 0)
      s[i]--;
  }
  int res = -1e9;
  for (int i = 1; i <= n; i++) {
    int l = lower_bound(range.begin(), range.end(), q[i].l) - range.begin();
    int r = upper_bound(range.begin(), range.end(), q[i].r) - range.begin() - 1;
    int ans = s[r] - s[max(l - 1, 0)];
    res = max(res, ans);
  }
  cout << res + dem + 1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) sol();
}

```