#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 7;
const int inf = 1e9 + 1;
int read() {
  int ans = 0, f = 1, c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + (c - '0');
    c = getchar();
  }
  return ans * f;
}
int T, n, ans, base, cnt;
struct node {
  int l, r;
} e[M];
bool cmp(node a, node b) { return a.l == b.l ? a.r < b.r : a.l < b.l; }
int main() {
  T = read();
  while (T--) {
    n = read();
    ans = 0;
    cnt = 0;
    base = 0;
    for (int i = 1; i <= n; i++) e[i].l = read(), e[i].r = read();
    sort(e + 1, e + 1 + n, cmp);
    bool f = 0;
    for (int i = 2; i <= n; i++)
      if (e[i].l <= e[i - 1].r) f = 1;
    if (!f) {
      printf("%d\n", n - 1);
      continue;
    }
    int mxr = -inf, L = inf, R = -inf, Tail = -inf;
    for (int i = 1; i <= n; i++) {
      if (mxr >= L && mxr < e[i].l && e[i].l <= R) cnt++;
      ans = max(ans, cnt);
      if (e[i].r < R)
        mxr = max(mxr, e[i].r);
      else {
        mxr = max(mxr, R);
        cnt = 0;
        L = e[i].l;
        R = e[i].r;
      }
      if (e[i].l > Tail) base++;
      Tail = max(Tail, e[i].r);
    }
    printf("%d\n", base + ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int M = 2e5 + 7;
const int inf = 1e9 + 1;
int read() {
  int ans = 0, f = 1, c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + (c - '0');
    c = getchar();
  }
  return ans * f;
}
int T, n, ans, base, cnt;
struct node {
  int l, r;
} e[M];
bool cmp(node a, node b) { return a.l == b.l ? a.r < b.r : a.l < b.l; }
int main() {
  T = read();
  while (T--) {
    n = read();
    ans = 0;
    cnt = 0;
    base = 0;
    for (int i = 1; i <= n; i++) e[i].l = read(), e[i].r = read();
    sort(e + 1, e + 1 + n, cmp);
    bool f = 0;
    for (int i = 2; i <= n; i++)
      if (e[i].l <= e[i - 1].r) f = 1;
    if (!f) {
      printf("%d\n", n - 1);
      continue;
    }
    int mxr = -inf, L = inf, R = -inf, Tail = -inf;
    for (int i = 1; i <= n; i++) {
      if (mxr >= L && mxr < e[i].l && e[i].l <= R) cnt++;
      ans = max(ans, cnt);
      if (e[i].r < R)
        mxr = max(mxr, e[i].r);
      else {
        mxr = max(mxr, R);
        cnt = 0;
        L = e[i].l;
        R = e[i].r;
      }
      if (e[i].l > Tail) base++;
      Tail = max(Tail, e[i].r);
    }
    printf("%d\n", base + ans);
  }
  return 0;
}

```