#include <bits/stdc++.h>
using namespace std;
int T, n, buf[400005], bn, L[400005], on, OL[400005], OR[400005], ans, cnt;
pair<int, int> P[200005];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    bn = on = ans = cnt = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &P[i].first, &P[i].second);
      P[i].first *= 2;
      P[i].second = P[i].second * 2 + 1;
      buf[++bn] = P[i].first, buf[++bn] = P[i].second;
    }
    sort(buf + 1, buf + bn + 1);
    bn = unique(buf + 1, buf + bn + 1) - buf - 1;
    for (int i = 1; i <= bn; i++) L[i] = 0;
    for (int i = 1; i <= n; i++) {
      P[i].first = lower_bound(buf + 1, buf + bn + 1, P[i].first) - buf;
      P[i].second = lower_bound(buf + 1, buf + bn + 1, P[i].second) - buf;
      L[P[i].first]++, L[P[i].second]--;
    }
    bool gg = false;
    for (int i = 1; i <= bn; i++) {
      L[i] += L[i - 1];
      if (L[i] > 1) gg = true;
    }
    if (!gg) {
      printf("%d\n", n - 1);
      continue;
    }
    for (int i = 1, j; i <= bn; i = j + 1) {
      j = i;
      while (j <= bn && L[j]) j++;
      if (j > i) cnt++;
    }
    for (int i = 1, j; i <= bn; i = j + 1) {
      j = i;
      while (j <= bn && L[j] == 1) j++;
      if (j > i && L[i - 1] && L[j]) ++on, OL[on] = i, OR[on] = j - 1;
    }
    for (int i = 1, l, r; i <= n; i++) {
      l = lower_bound(OR + 1, OR + on + 1, P[i].first) - OR;
      r = upper_bound(OL + 1, OL + on + 1, P[i].second - 1) - OL;
      ans = max(ans, cnt + r - l);
    }
    printf("%d\n", ans);
  }
  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>
using namespace std;
int T, n, buf[400005], bn, L[400005], on, OL[400005], OR[400005], ans, cnt;
pair<int, int> P[200005];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    bn = on = ans = cnt = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &P[i].first, &P[i].second);
      P[i].first *= 2;
      P[i].second = P[i].second * 2 + 1;
      buf[++bn] = P[i].first, buf[++bn] = P[i].second;
    }
    sort(buf + 1, buf + bn + 1);
    bn = unique(buf + 1, buf + bn + 1) - buf - 1;
    for (int i = 1; i <= bn; i++) L[i] = 0;
    for (int i = 1; i <= n; i++) {
      P[i].first = lower_bound(buf + 1, buf + bn + 1, P[i].first) - buf;
      P[i].second = lower_bound(buf + 1, buf + bn + 1, P[i].second) - buf;
      L[P[i].first]++, L[P[i].second]--;
    }
    bool gg = false;
    for (int i = 1; i <= bn; i++) {
      L[i] += L[i - 1];
      if (L[i] > 1) gg = true;
    }
    if (!gg) {
      printf("%d\n", n - 1);
      continue;
    }
    for (int i = 1, j; i <= bn; i = j + 1) {
      j = i;
      while (j <= bn && L[j]) j++;
      if (j > i) cnt++;
    }
    for (int i = 1, j; i <= bn; i = j + 1) {
      j = i;
      while (j <= bn && L[j] == 1) j++;
      if (j > i && L[i - 1] && L[j]) ++on, OL[on] = i, OR[on] = j - 1;
    }
    for (int i = 1, l, r; i <= n; i++) {
      l = lower_bound(OR + 1, OR + on + 1, P[i].first) - OR;
      r = upper_bound(OL + 1, OL + on + 1, P[i].second - 1) - OL;
      ans = max(ans, cnt + r - l);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```