#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return w == 1 ? x : -x;
}
int n, cnt, tot, l[800005], r[800005], a[800005 << 1], lx[800005], rx[800005],
    p[800005], t[800005], p2[800005], h[800005];
int main() {
  int T = read();
  while (T--) {
    n = read();
    cnt = 0;
    tot = 0;
    for (int i = 1; i <= n; i++) {
      l[i] = read(), r[i] = read();
      a[++cnt] = l[i], a[++cnt] = r[i];
    }
    sort(a + 1, a + cnt + 1);
    tot = unique(a + 1, a + cnt + 1) - a - 1;
    for (int i = 1; i <= n; i++) {
      lx[i] = lower_bound(a + 1, a + tot + 1, l[i]) - a;
      rx[i] = lower_bound(a + 1, a + tot + 1, r[i]) - a;
      lx[i] <<= 1;
      rx[i] <<= 1;
    }
    int tmp = 0;
    for (int i = 1; i <= tot * 2; i++) p2[i] = 0, h[i] = 0, p[i] = 0;
    for (int i = 1; i <= tot * 2; i++) t[i] = 0;
    for (int i = 1; i <= n; i++) t[lx[i]]++, t[rx[i] + 1]--;
    for (int i = 1; i <= tot * 2; i++) tmp += t[i], p[i] = tmp;
    for (int i = 1; i <= tot * 2; i++)
      if (p[i] != p[i - 1] && p[i] == 1) p2[i] = 1;
    for (int i = 1; i <= tot * 2; i++) h[i] = h[i - 1] + p2[i];
    int ans = -1e9, ans1 = 0;
    for (int i = 1; i <= tot * 2 + 1; i++)
      if (p[i] != 0 && p[i - 1] == 0) ans1++;
    for (int i = 1; i <= n; i++) {
      int q = h[rx[i]] - h[lx[i] - 1];
      q -= (p[rx[i]] == 1), q -= (p[lx[i]] == 1);
      ans = max(ans, q);
    }
    cout << ans1 + ans << endl;
  }
  return 0;
}

### 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(2)
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return w == 1 ? x : -x;
}
int n, cnt, tot, l[800005], r[800005], a[800005 << 1], lx[800005], rx[800005],
    p[800005], t[800005], p2[800005], h[800005];
int main() {
  int T = read();
  while (T--) {
    n = read();
    cnt = 0;
    tot = 0;
    for (int i = 1; i <= n; i++) {
      l[i] = read(), r[i] = read();
      a[++cnt] = l[i], a[++cnt] = r[i];
    }
    sort(a + 1, a + cnt + 1);
    tot = unique(a + 1, a + cnt + 1) - a - 1;
    for (int i = 1; i <= n; i++) {
      lx[i] = lower_bound(a + 1, a + tot + 1, l[i]) - a;
      rx[i] = lower_bound(a + 1, a + tot + 1, r[i]) - a;
      lx[i] <<= 1;
      rx[i] <<= 1;
    }
    int tmp = 0;
    for (int i = 1; i <= tot * 2; i++) p2[i] = 0, h[i] = 0, p[i] = 0;
    for (int i = 1; i <= tot * 2; i++) t[i] = 0;
    for (int i = 1; i <= n; i++) t[lx[i]]++, t[rx[i] + 1]--;
    for (int i = 1; i <= tot * 2; i++) tmp += t[i], p[i] = tmp;
    for (int i = 1; i <= tot * 2; i++)
      if (p[i] != p[i - 1] && p[i] == 1) p2[i] = 1;
    for (int i = 1; i <= tot * 2; i++) h[i] = h[i - 1] + p2[i];
    int ans = -1e9, ans1 = 0;
    for (int i = 1; i <= tot * 2 + 1; i++)
      if (p[i] != 0 && p[i - 1] == 0) ans1++;
    for (int i = 1; i <= n; i++) {
      int q = h[rx[i]] - h[lx[i] - 1];
      q -= (p[rx[i]] == 1), q -= (p[lx[i]] == 1);
      ans = max(ans, q);
    }
    cout << ans1 + ans << endl;
  }
  return 0;
}

```