#include <bits/stdc++.h>
using namespace std;
struct Node {
  int a, b;
} p[200005];
Node s[200005];
int tail = 0;
int n, mina[200005], ans;
bool cmp(Node nxt, Node pre) {
  if (nxt.b != pre.b) return nxt.b < pre.b;
  return nxt.a > pre.a;
}
void calc(int &x, int y, int cf, int flag) {
  int num = 0;
  while (tail > 0) {
    Node t = s[tail];
    if (t.b < x) break;
    if (t.b <= y) num++;
    x = min(x, t.a);
    tail--;
  }
  if (cf) ans = max(ans, num - flag);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    ans = -99999999;
    tail = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d%d", &p[i].a, &p[i].b);
    sort(p + 1, p + 1 + n, cmp);
    mina[n] = 1999999999;
    for (int i = n - 1; i >= 1; i--) mina[i] = min(p[i + 1].a, mina[i + 1]);
    s[++tail] = p[1];
    for (int i = 2; i <= n; i++) {
      calc(p[i].a, min(p[i].b, mina[i] - 1), mina[i] > p[i].a,
           mina[i] > p[i].b);
      s[++tail] = p[i];
    }
    ans = ans + tail;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
struct Node {
  int a, b;
} p[200005];
Node s[200005];
int tail = 0;
int n, mina[200005], ans;
bool cmp(Node nxt, Node pre) {
  if (nxt.b != pre.b) return nxt.b < pre.b;
  return nxt.a > pre.a;
}
void calc(int &x, int y, int cf, int flag) {
  int num = 0;
  while (tail > 0) {
    Node t = s[tail];
    if (t.b < x) break;
    if (t.b <= y) num++;
    x = min(x, t.a);
    tail--;
  }
  if (cf) ans = max(ans, num - flag);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    ans = -99999999;
    tail = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d%d", &p[i].a, &p[i].b);
    sort(p + 1, p + 1 + n, cmp);
    mina[n] = 1999999999;
    for (int i = n - 1; i >= 1; i--) mina[i] = min(p[i + 1].a, mina[i + 1]);
    s[++tail] = p[1];
    for (int i = 2; i <= n; i++) {
      calc(p[i].a, min(p[i].b, mina[i] - 1), mina[i] > p[i].a,
           mina[i] > p[i].b);
      s[++tail] = p[i];
    }
    ans = ans + tail;
    printf("%d\n", ans);
  }
  return 0;
}

```