#include <bits/stdc++.h>
using namespace std;
struct seg {
  int orix, x, num, state;
} a[400001];
bool cmp1(seg p, seg q) {
  return p.orix == q.orix ? p.state < q.state : p.orix < q.orix;
}
int te, n, tsize, sta[200001], en[200001];
long long uni, ans, now;
long long tree1[4000001], lazy[4000001], tree2[4000001];
void update1(int p, int q, int s, int e, int cur) {
  if (e < p || q < s)
    return;
  else if (p <= s && e <= q)
    lazy[cur]++;
  else {
    if (lazy[cur]) {
      tree1[cur] += lazy[cur] * (e - s + 1);
      if (cur < tsize) {
        lazy[2 * cur] += lazy[cur];
        lazy[2 * cur + 1] += lazy[cur];
      }
      lazy[cur] = 0;
    }
    int mid = (s + e) / 2;
    update1(p, q, s, mid, 2 * cur);
    update1(p, q, mid + 1, e, 2 * cur + 1);
  }
}
long long sum1(int p, int q, int s, int e, int cur) {
  if (lazy[cur]) {
    tree1[cur] += lazy[cur] * (e - s + 1);
    if (cur < tsize) {
      lazy[2 * cur] += lazy[cur];
      lazy[2 * cur + 1] += lazy[cur];
    }
    lazy[cur] = 0;
  }
  if (e < p || q < s) return 0;
  if (p <= s && e <= q) return tree1[cur];
  int mid = (s + e) / 2;
  return sum1(p, q, s, mid, 2 * cur) + sum1(p, q, mid + 1, e, 2 * cur + 1);
}
void update2(int p) {
  int cur = tsize + p - 1;
  while (cur) {
    tree2[cur]++;
    cur /= 2;
  }
}
long long sum2(int p, int q, int s, int e, int cur) {
  if (e < p || q < s) return 0;
  if (p <= s && e <= q) return tree2[cur];
  int mid = (s + e) / 2;
  return sum2(p, q, s, mid, 2 * cur) + sum2(p, q, mid + 1, e, 2 * cur + 1);
}
int main() {
  scanf("%d", &te);
  while (te--) {
    scanf("%d", &n);
    for (int p, q, i = 1; i <= n; i++) {
      scanf("%d%d", &p, &q);
      a[2 * i - 1] = {p, 0, i, 1};
      a[2 * i] = {q, 0, i, 2};
    }
    sort(a + 1, a + 2 * n + 1, cmp1);
    a[0] = {-1000000001, 0, 0, 0};
    ans = -1;
    uni = now = 0;
    for (int i = 1; i <= 2 * n; i++) {
      if (a[i].state == 1) {
        if (now == 0) uni++;
        now++;
      } else
        now--;
      if (a[i - 1].orix == a[i].orix)
        a[i].x = a[i - 1].x;
      else
        a[i].x = a[i - 1].x + 1;
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (a[i].state == 1)
        sta[a[i].num] = a[i].x * 2;
      else
        en[a[i].num] = a[i].x * 2;
    }
    for (tsize = 1; tsize < 2 * a[2 * n].x; tsize *= 2)
      ;
    for (int i = 2 * tsize - 1; i > 0; i--) tree1[i] = tree2[i] = lazy[i] = 0;
    for (int i = 1; i <= n; i++) update1(sta[i], en[i], 1, tsize, 1);
    for (int i = 1; i <= tsize; i++) {
      long long now = sum1(i, i, 1, tsize, 1);
      if (now == 1) update2(i);
    }
    for (int i = 1; i <= n; i++) {
      if (sum2(sta[i], en[i], 1, tsize, 1) == en[i] - sta[i] + 1)
        continue;
      else
        ans =
            max(ans, sum2(sta[i] + 1, en[i] - 1, 1, tsize, 1) -
                         tree2[sta[i] + tsize - 1] - tree2[en[i] + tsize - 1]);
    }
    printf("%lld\n", ans + uni);
  }
}

### Prompt

Create a solution in CPP for 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;
struct seg {
  int orix, x, num, state;
} a[400001];
bool cmp1(seg p, seg q) {
  return p.orix == q.orix ? p.state < q.state : p.orix < q.orix;
}
int te, n, tsize, sta[200001], en[200001];
long long uni, ans, now;
long long tree1[4000001], lazy[4000001], tree2[4000001];
void update1(int p, int q, int s, int e, int cur) {
  if (e < p || q < s)
    return;
  else if (p <= s && e <= q)
    lazy[cur]++;
  else {
    if (lazy[cur]) {
      tree1[cur] += lazy[cur] * (e - s + 1);
      if (cur < tsize) {
        lazy[2 * cur] += lazy[cur];
        lazy[2 * cur + 1] += lazy[cur];
      }
      lazy[cur] = 0;
    }
    int mid = (s + e) / 2;
    update1(p, q, s, mid, 2 * cur);
    update1(p, q, mid + 1, e, 2 * cur + 1);
  }
}
long long sum1(int p, int q, int s, int e, int cur) {
  if (lazy[cur]) {
    tree1[cur] += lazy[cur] * (e - s + 1);
    if (cur < tsize) {
      lazy[2 * cur] += lazy[cur];
      lazy[2 * cur + 1] += lazy[cur];
    }
    lazy[cur] = 0;
  }
  if (e < p || q < s) return 0;
  if (p <= s && e <= q) return tree1[cur];
  int mid = (s + e) / 2;
  return sum1(p, q, s, mid, 2 * cur) + sum1(p, q, mid + 1, e, 2 * cur + 1);
}
void update2(int p) {
  int cur = tsize + p - 1;
  while (cur) {
    tree2[cur]++;
    cur /= 2;
  }
}
long long sum2(int p, int q, int s, int e, int cur) {
  if (e < p || q < s) return 0;
  if (p <= s && e <= q) return tree2[cur];
  int mid = (s + e) / 2;
  return sum2(p, q, s, mid, 2 * cur) + sum2(p, q, mid + 1, e, 2 * cur + 1);
}
int main() {
  scanf("%d", &te);
  while (te--) {
    scanf("%d", &n);
    for (int p, q, i = 1; i <= n; i++) {
      scanf("%d%d", &p, &q);
      a[2 * i - 1] = {p, 0, i, 1};
      a[2 * i] = {q, 0, i, 2};
    }
    sort(a + 1, a + 2 * n + 1, cmp1);
    a[0] = {-1000000001, 0, 0, 0};
    ans = -1;
    uni = now = 0;
    for (int i = 1; i <= 2 * n; i++) {
      if (a[i].state == 1) {
        if (now == 0) uni++;
        now++;
      } else
        now--;
      if (a[i - 1].orix == a[i].orix)
        a[i].x = a[i - 1].x;
      else
        a[i].x = a[i - 1].x + 1;
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (a[i].state == 1)
        sta[a[i].num] = a[i].x * 2;
      else
        en[a[i].num] = a[i].x * 2;
    }
    for (tsize = 1; tsize < 2 * a[2 * n].x; tsize *= 2)
      ;
    for (int i = 2 * tsize - 1; i > 0; i--) tree1[i] = tree2[i] = lazy[i] = 0;
    for (int i = 1; i <= n; i++) update1(sta[i], en[i], 1, tsize, 1);
    for (int i = 1; i <= tsize; i++) {
      long long now = sum1(i, i, 1, tsize, 1);
      if (now == 1) update2(i);
    }
    for (int i = 1; i <= n; i++) {
      if (sum2(sta[i], en[i], 1, tsize, 1) == en[i] - sta[i] + 1)
        continue;
      else
        ans =
            max(ans, sum2(sta[i] + 1, en[i] - 1, 1, tsize, 1) -
                         tree2[sta[i] + tsize - 1] - tree2[en[i] + tsize - 1]);
    }
    printf("%lld\n", ans + uni);
  }
}

```