#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 998244353, M = 1000000;
long long n, m, t, k, k2, f, mi = 2e18, x, y, z, ma, x3, x2, y2, l, r, i, j,
                             ans, b[410000], d[410000], de[4010000], p[410000];
string s, s2;
vector<long long> v[1100000], v2[1100000], v3;
map<long long, long long> mp, mp2;
map<long long, long long>::iterator it;
struct pos {
  long long x, y;
};
pos a[410000];
bool cmp(pos l, pos r) {
  if (l.x == r.x) return l.y < r.y;
  return l.x < r.x;
}
void bild(int l, int r, int ve) {
  if (l == r) {
    de[ve] = (d[l] == 1);
    return;
  }
  int t = (l + r) / 2;
  bild(l, t, ve * 2);
  bild(t + 1, r, ve * 2 + 1);
  de[ve] = de[ve * 2] + de[ve * 2 + 1];
}
int sum(int l, int r, int x, int y, int ve) {
  if (r < x || l > y) return 0;
  if (l >= x && r <= y) return de[ve];
  int t = (l + r) / 2;
  return sum(l, t, x, y, ve * 2) + sum(t + 1, r, x, y, ve * 2 + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    cin >> n;
    mp = mp2;
    for (i = 1; i <= n; i++) {
      cin >> a[i].x >> a[i].y;
      mp[a[i].x] = 1;
      mp[a[i].y] = 1;
    }
    m = 0;
    for (it = mp.begin(); it != mp.end(); it++) {
      int x = it->first;
      m++;
      mp[x] = m;
    }
    for (i = 1; i <= n; i++) {
      a[i].x = mp[a[i].x];
      a[i].y = mp[a[i].y];
      v[a[i].x].push_back(a[i].y);
      v2[a[i].y].push_back(a[i].x);
      a[i].y--;
    }
    ans = 0;
    f = 0;
    t = 0;
    for (i = 1; i <= m; i++) {
      t += (int)v[i].size();
      t -= (int)v2[i].size();
      d[i] = t;
      if (t == 0) ans++;
    }
    m--;
    m = max(m, 1ll);
    bild(1, m, 1);
    ma = -1e9;
    for (i = 1; i <= n; i++) {
      if (a[i].y - a[i].x < 0) {
        if ((int)v[a[i].x].size() + (int)v2[a[i].x].size() == 2)
          ma = max(ma, ans - 1);
        else
          ma = max(ma, ans);
        continue;
      }
      k = sum(1, m, a[i].x, a[i].y, 1) -
          ((int)v[a[i].x].size() + (int)v2[a[i].x].size() == 1 &&
           d[a[i].x] == 1) -
          ((int)v[a[i].y + 1].size() + (int)v2[a[i].y + 1].size() == 1 &&
           d[a[i].y] == 1);
      ma = max(ma, ans + k);
    }
    cout << ma << endl;
    for (i = 1; i <= 4 * m + 1; i++) de[i] = 0;
    for (i = 1; i <= m + 1; i++) v[i] = v3, v2[i] = v3;
  }
}

### Prompt

Generate 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 998244353, M = 1000000;
long long n, m, t, k, k2, f, mi = 2e18, x, y, z, ma, x3, x2, y2, l, r, i, j,
                             ans, b[410000], d[410000], de[4010000], p[410000];
string s, s2;
vector<long long> v[1100000], v2[1100000], v3;
map<long long, long long> mp, mp2;
map<long long, long long>::iterator it;
struct pos {
  long long x, y;
};
pos a[410000];
bool cmp(pos l, pos r) {
  if (l.x == r.x) return l.y < r.y;
  return l.x < r.x;
}
void bild(int l, int r, int ve) {
  if (l == r) {
    de[ve] = (d[l] == 1);
    return;
  }
  int t = (l + r) / 2;
  bild(l, t, ve * 2);
  bild(t + 1, r, ve * 2 + 1);
  de[ve] = de[ve * 2] + de[ve * 2 + 1];
}
int sum(int l, int r, int x, int y, int ve) {
  if (r < x || l > y) return 0;
  if (l >= x && r <= y) return de[ve];
  int t = (l + r) / 2;
  return sum(l, t, x, y, ve * 2) + sum(t + 1, r, x, y, ve * 2 + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    cin >> n;
    mp = mp2;
    for (i = 1; i <= n; i++) {
      cin >> a[i].x >> a[i].y;
      mp[a[i].x] = 1;
      mp[a[i].y] = 1;
    }
    m = 0;
    for (it = mp.begin(); it != mp.end(); it++) {
      int x = it->first;
      m++;
      mp[x] = m;
    }
    for (i = 1; i <= n; i++) {
      a[i].x = mp[a[i].x];
      a[i].y = mp[a[i].y];
      v[a[i].x].push_back(a[i].y);
      v2[a[i].y].push_back(a[i].x);
      a[i].y--;
    }
    ans = 0;
    f = 0;
    t = 0;
    for (i = 1; i <= m; i++) {
      t += (int)v[i].size();
      t -= (int)v2[i].size();
      d[i] = t;
      if (t == 0) ans++;
    }
    m--;
    m = max(m, 1ll);
    bild(1, m, 1);
    ma = -1e9;
    for (i = 1; i <= n; i++) {
      if (a[i].y - a[i].x < 0) {
        if ((int)v[a[i].x].size() + (int)v2[a[i].x].size() == 2)
          ma = max(ma, ans - 1);
        else
          ma = max(ma, ans);
        continue;
      }
      k = sum(1, m, a[i].x, a[i].y, 1) -
          ((int)v[a[i].x].size() + (int)v2[a[i].x].size() == 1 &&
           d[a[i].x] == 1) -
          ((int)v[a[i].y + 1].size() + (int)v2[a[i].y + 1].size() == 1 &&
           d[a[i].y] == 1);
      ma = max(ma, ans + k);
    }
    cout << ma << endl;
    for (i = 1; i <= 4 * m + 1; i++) de[i] = 0;
    for (i = 1; i <= m + 1; i++) v[i] = v3, v2[i] = v3;
  }
}

```