#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T read(T &r) {
  static char c;
  static int flag;
  flag = 1, r = 0;
  for (c = getchar(); c > '9' || c < '0'; c = getchar())
    if (c == '-') flag = -1;
  for (; c >= '0' && c <= '9';
       r = (r << 1) + (r << 3) + (c ^ 48), c = getchar())
    ;
  return r *= flag;
}
const int maxn = 1e6 + 100;
struct node {
  int l, r;
} a[maxn];
int X[maxn], cnt;
int T;
int n;
inline int search(int x) {
  int ll = 1, rr = cnt, mid, ans;
  while (ll <= rr) {
    mid = (ll + rr) >> 1;
    if (X[mid] <= x) {
      ans = mid;
      ll = mid + 1;
    } else
      rr = mid - 1;
  }
  return ans;
}
int A[maxn];
int B[maxn];
int sum[maxn];
int main() {
  read(T);
  while (T--) {
    cnt = 0;
    read(n);
    for (int i = 1; i <= n; ++i)
      read(a[i].l), read(a[i].r), X[++cnt] = a[i].l, X[++cnt] = a[i].r;
    sort(X + 1, X + 1 + cnt);
    cnt = unique(X + 1, X + 1 + cnt) - X - 1;
    for (int i = 1; i <= n; ++i) {
      a[i].l = search(a[i].l);
      a[i].r = search(a[i].r);
      a[i].l <<= 1;
      a[i].r <<= 1;
      A[a[i].l]++;
      A[a[i].r + 1]--;
    }
    cnt <<= 1;
    cnt += 10;
    for (int i = 1; i <= cnt; ++i) A[i] += A[i - 1];
    int origin = 0;
    for (int i = 0; i <= cnt; ++i) origin += (A[i] && !A[i + 1]);
    for (int i = 0; i < cnt; ++i)
      if (A[i] != 1 && A[i + 1] == 1) B[i + 1]++;
    for (int i = 1; i <= cnt; ++i)
      if (A[i] == 1 && A[i + 1] != 1) B[i]++;
    for (int i = 1; i <= cnt; ++i) B[i] += B[i - 1];
    int maxx = -(1 << 30);
    for (int i = 1; i <= n; ++i) {
      int temp = (B[a[i].r] - B[a[i].l - 1]) >> 1;
      temp -= (A[a[i].l] == 1) + (A[a[i].r] == 1);
      if (maxx < temp) maxx = temp;
    }
    printf("%d\n", origin + maxx);
    for (int i = 0; i <= cnt; ++i) A[i] = B[i] = 0;
  }
  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;
template <typename T>
inline T read(T &r) {
  static char c;
  static int flag;
  flag = 1, r = 0;
  for (c = getchar(); c > '9' || c < '0'; c = getchar())
    if (c == '-') flag = -1;
  for (; c >= '0' && c <= '9';
       r = (r << 1) + (r << 3) + (c ^ 48), c = getchar())
    ;
  return r *= flag;
}
const int maxn = 1e6 + 100;
struct node {
  int l, r;
} a[maxn];
int X[maxn], cnt;
int T;
int n;
inline int search(int x) {
  int ll = 1, rr = cnt, mid, ans;
  while (ll <= rr) {
    mid = (ll + rr) >> 1;
    if (X[mid] <= x) {
      ans = mid;
      ll = mid + 1;
    } else
      rr = mid - 1;
  }
  return ans;
}
int A[maxn];
int B[maxn];
int sum[maxn];
int main() {
  read(T);
  while (T--) {
    cnt = 0;
    read(n);
    for (int i = 1; i <= n; ++i)
      read(a[i].l), read(a[i].r), X[++cnt] = a[i].l, X[++cnt] = a[i].r;
    sort(X + 1, X + 1 + cnt);
    cnt = unique(X + 1, X + 1 + cnt) - X - 1;
    for (int i = 1; i <= n; ++i) {
      a[i].l = search(a[i].l);
      a[i].r = search(a[i].r);
      a[i].l <<= 1;
      a[i].r <<= 1;
      A[a[i].l]++;
      A[a[i].r + 1]--;
    }
    cnt <<= 1;
    cnt += 10;
    for (int i = 1; i <= cnt; ++i) A[i] += A[i - 1];
    int origin = 0;
    for (int i = 0; i <= cnt; ++i) origin += (A[i] && !A[i + 1]);
    for (int i = 0; i < cnt; ++i)
      if (A[i] != 1 && A[i + 1] == 1) B[i + 1]++;
    for (int i = 1; i <= cnt; ++i)
      if (A[i] == 1 && A[i + 1] != 1) B[i]++;
    for (int i = 1; i <= cnt; ++i) B[i] += B[i - 1];
    int maxx = -(1 << 30);
    for (int i = 1; i <= n; ++i) {
      int temp = (B[a[i].r] - B[a[i].l - 1]) >> 1;
      temp -= (A[a[i].l] == 1) + (A[a[i].r] == 1);
      if (maxx < temp) maxx = temp;
    }
    printf("%d\n", origin + maxx);
    for (int i = 0; i <= cnt; ++i) A[i] = B[i] = 0;
  }
  return 0;
}

```