#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 7;
int n, m;
struct segment {
  int L, R;
} a[maxn];
bool cmp1(const segment &x, const segment &y) { return x.L < y.L; }
struct border {
  int p, id, flag;
} b[2 * maxn];
bool cmp2(const border &x, const border &y) { return x.p < y.p; }
map<int, int> Lb;
int init() {
  int cnt = 0;
  Lb.clear();
  int L, R = -INF;
  for (int i = 1; i <= n; i++) {
    if (a[i].L > R) {
      if (R != -INF) Lb[L] = 0;
      cnt++;
      L = a[i].L, R = a[i].R;
    } else
      R = max(R, a[i].R);
  }
  Lb[L] = 0;
  return cnt;
}
int ans[maxn];
void process() {
  set<int> open;
  for (int i = 1; i <= m; i++) {
    vector<int> op, cl;
    int j = i;
    do {
      if (b[j].flag == -1)
        op.push_back(b[j].id);
      else
        cl.push_back(b[j].id);
      j++;
    } while (j <= m && b[j].p == b[i].p);
    if (open.size() == 1 && !op.empty()) ans[*open.begin()]++;
    for (auto it : op) open.insert(it);
    for (auto it : cl) open.erase(it);
    i = j - 1;
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d %d", &a[i].L, &a[i].R);
    sort(a + 1, a + n + 1, cmp1);
    int cur = init();
    m = 0;
    for (int i = 1; i <= n; i++) {
      b[++m] = border{a[i].L, i, -1};
      b[++m] = border{a[i].R, i, 1};
    }
    sort(b + 1, b + m + 1, cmp2);
    memset(ans, 0, (n + 1) * sizeof(int));
    process();
    for (int i = 1; i <= n; i++) {
      if (Lb.count(a[i].L)) Lb[a[i].L]++;
    }
    for (int i = 1; i <= n; i++) {
      if (Lb[a[i].L] == 1) ans[i]--;
    }
    printf("%d\n", *max_element(ans + 1, ans + n + 1) + cur);
  }
  return 0;
}

### Prompt

Please formulate 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;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 7;
int n, m;
struct segment {
  int L, R;
} a[maxn];
bool cmp1(const segment &x, const segment &y) { return x.L < y.L; }
struct border {
  int p, id, flag;
} b[2 * maxn];
bool cmp2(const border &x, const border &y) { return x.p < y.p; }
map<int, int> Lb;
int init() {
  int cnt = 0;
  Lb.clear();
  int L, R = -INF;
  for (int i = 1; i <= n; i++) {
    if (a[i].L > R) {
      if (R != -INF) Lb[L] = 0;
      cnt++;
      L = a[i].L, R = a[i].R;
    } else
      R = max(R, a[i].R);
  }
  Lb[L] = 0;
  return cnt;
}
int ans[maxn];
void process() {
  set<int> open;
  for (int i = 1; i <= m; i++) {
    vector<int> op, cl;
    int j = i;
    do {
      if (b[j].flag == -1)
        op.push_back(b[j].id);
      else
        cl.push_back(b[j].id);
      j++;
    } while (j <= m && b[j].p == b[i].p);
    if (open.size() == 1 && !op.empty()) ans[*open.begin()]++;
    for (auto it : op) open.insert(it);
    for (auto it : cl) open.erase(it);
    i = j - 1;
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d %d", &a[i].L, &a[i].R);
    sort(a + 1, a + n + 1, cmp1);
    int cur = init();
    m = 0;
    for (int i = 1; i <= n; i++) {
      b[++m] = border{a[i].L, i, -1};
      b[++m] = border{a[i].R, i, 1};
    }
    sort(b + 1, b + m + 1, cmp2);
    memset(ans, 0, (n + 1) * sizeof(int));
    process();
    for (int i = 1; i <= n; i++) {
      if (Lb.count(a[i].L)) Lb[a[i].L]++;
    }
    for (int i = 1; i <= n; i++) {
      if (Lb[a[i].L] == 1) ans[i]--;
    }
    printf("%d\n", *max_element(ans + 1, ans + n + 1) + cur);
  }
  return 0;
}

```