#include <bits/stdc++.h>
using namespace std;
const int N = 8e5 + 5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(int x = -1, char c = '\n') {
  write(x);
  putchar(c);
}
int l[N], r[N], b[N];
int c[N], a[N], one[N];
signed main(signed Recall, char *_902_[]) {
  (void)Recall, (void)_902_;
  int T = read();
  while (T--) {
    int n = read(), cnt = 0, ans = -1, atfirst = 0, minl = 1919810111, maxr = 0;
    for (int i = 1; i <= n; i++)
      b[++cnt] = l[i] = read(), b[++cnt] = r[i] = read();
    sort(b + 1, b + 1 + cnt);
    cnt = unique(b + 1, b + 1 + cnt) - b - 1;
    for (int i = 1; i <= n; i++)
      l[i] = lower_bound(b + 1, b + 1 + cnt, l[i]) - b,
      r[i] = lower_bound(b + 1, b + 1 + cnt, r[i]) - b;
    for (int i = 1; i <= n; i++) l[i] = l[i] * 2 - 1, r[i] = r[i] * 2 - 1;
    for (int i = 1; i <= n; i++) minl = min(minl, l[i]), maxr = max(maxr, r[i]);
    cnt = cnt * 2 + 1;
    for (int i = 1; i <= n; i++) c[l[i]]++, c[r[i] + 1]--;
    for (int i = 1; i <= cnt; i++) c[i] += c[i - 1];
    for (int i = 1; i <= cnt; i++)
      if (c[i] == 1 && c[i - 1] > 1) a[i] = 1;
    for (int i = 1; i <= cnt; i++)
      if (c[i] == 1) one[i] = 1;
    for (int i = 1; i <= cnt; i++) one[i] += one[i - 1], a[i] += a[i - 1];
    for (int i = 1; i <= cnt; i++) atfirst += c[i] && !c[i - 1];
    for (int i = 1; i <= n; i++) {
      int l = ::l[i], r = ::r[i];
      if (one[r] - one[l - 1] == r - l + 1) {
        int now = !!c[l - 1] && !!c[r + 1];
        if (!c[l - 1] && !c[r + 1]) now = -1;
        ans = max(ans, now);
        continue;
      }
      int now = a[r] - a[l - 1];
      if (c[r] == 1 && (!c[r + 1] || r == maxr)) now--;
      if (c[l] == 1 && c[l - 1] && l > minl) now++;
      ans = max(ans, now);
    }
    print(ans + atfirst);
    for (int i = 1; i <= cnt; i++) c[i] = a[i] = one[i] = 0;
  }
  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;
const int N = 8e5 + 5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(int x = -1, char c = '\n') {
  write(x);
  putchar(c);
}
int l[N], r[N], b[N];
int c[N], a[N], one[N];
signed main(signed Recall, char *_902_[]) {
  (void)Recall, (void)_902_;
  int T = read();
  while (T--) {
    int n = read(), cnt = 0, ans = -1, atfirst = 0, minl = 1919810111, maxr = 0;
    for (int i = 1; i <= n; i++)
      b[++cnt] = l[i] = read(), b[++cnt] = r[i] = read();
    sort(b + 1, b + 1 + cnt);
    cnt = unique(b + 1, b + 1 + cnt) - b - 1;
    for (int i = 1; i <= n; i++)
      l[i] = lower_bound(b + 1, b + 1 + cnt, l[i]) - b,
      r[i] = lower_bound(b + 1, b + 1 + cnt, r[i]) - b;
    for (int i = 1; i <= n; i++) l[i] = l[i] * 2 - 1, r[i] = r[i] * 2 - 1;
    for (int i = 1; i <= n; i++) minl = min(minl, l[i]), maxr = max(maxr, r[i]);
    cnt = cnt * 2 + 1;
    for (int i = 1; i <= n; i++) c[l[i]]++, c[r[i] + 1]--;
    for (int i = 1; i <= cnt; i++) c[i] += c[i - 1];
    for (int i = 1; i <= cnt; i++)
      if (c[i] == 1 && c[i - 1] > 1) a[i] = 1;
    for (int i = 1; i <= cnt; i++)
      if (c[i] == 1) one[i] = 1;
    for (int i = 1; i <= cnt; i++) one[i] += one[i - 1], a[i] += a[i - 1];
    for (int i = 1; i <= cnt; i++) atfirst += c[i] && !c[i - 1];
    for (int i = 1; i <= n; i++) {
      int l = ::l[i], r = ::r[i];
      if (one[r] - one[l - 1] == r - l + 1) {
        int now = !!c[l - 1] && !!c[r + 1];
        if (!c[l - 1] && !c[r + 1]) now = -1;
        ans = max(ans, now);
        continue;
      }
      int now = a[r] - a[l - 1];
      if (c[r] == 1 && (!c[r + 1] || r == maxr)) now--;
      if (c[l] == 1 && c[l - 1] && l > minl) now++;
      ans = max(ans, now);
    }
    print(ans + atfirst);
    for (int i = 1; i <= cnt; i++) c[i] = a[i] = one[i] = 0;
  }
  return 0;
}

```