#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n;
int s[MAXN], e[MAXN];
int cnt, tot;
struct unq {
  int val, id, op;
} u[4 * MAXN];
struct tr {
  int cov, l, r, sum;
} t[16 * MAXN], an;
int ans;
bool cmp(const unq &a, const unq &b) { return a.val < b.val; }
tr operator+(const tr &a, const tr &b) {
  if (a.r == b.l && a.r) {
    return (tr){0, a.l, b.r, a.sum + b.sum - 2};
  }
  return (tr){0, a.l, b.r, a.sum + b.sum};
}
void update(int l, int r, int z, int y, int v, int id) {
  if (l == z && r == y) {
    t[id].cov += v;
    if (!t[id].cov) {
      if (l == r) {
        t[id].l = t[id].r = 0;
        t[id].sum = 0;
      } else
        t[id] = t[id * 2] + t[id * 2 + 1];
    } else
      t[id].r = t[id].l = 1, t[id].sum = 2;
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= y)
    update(l, mid, z, y, v, id * 2);
  else if (mid < z)
    update(mid + 1, r, z, y, v, id * 2 + 1);
  else {
    update(l, mid, z, mid, v, id * 2);
    update(mid + 1, r, mid + 1, y, v, id * 2 + 1);
  }
  if (!t[id].cov) t[id] = t[id * 2] + t[id * 2 + 1];
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    ans = cnt = tot = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &s[i], &e[i]);
      s[i] *= 2;
      e[i] *= 2;
      cnt++;
      u[cnt].val = s[i];
      u[cnt].id = i;
      u[cnt].op = 1;
      cnt++;
      u[cnt].val = s[i] - 1;
      u[cnt].id = i;
      u[cnt].op = 3;
      cnt++;
      u[cnt].val = e[i];
      u[cnt].id = i;
      u[cnt].op = 2;
      cnt++;
      u[cnt].val = e[i] + 1;
      u[cnt].id = i;
      u[cnt].op = 3;
    }
    sort(u + 1, u + 1 + cnt, cmp);
    for (int i = 1; i <= cnt; i++) {
      if (i == 1 || u[i].val != u[i - 1].val) tot++;
      if (u[i].op == 1) s[u[i].id] = tot;
      if (u[i].op == 2) e[u[i].id] = tot;
    }
    for (int i = 1; i <= n; i++) {
      update(1, tot, s[i], e[i], 1, 1);
    }
    for (int i = 1; i <= n; i++) {
      update(1, tot, s[i], e[i], -1, 1);
      ans = max(ans, t[1].sum - 2);
      update(1, tot, s[i], e[i], 1, 1);
    }
    printf("%d\n", ans / 2 + 1);
    for (int i = 1; i <= n; i++) update(1, tot, s[i], e[i], -1, 1);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MAXN = 2e5 + 5;
int n;
int s[MAXN], e[MAXN];
int cnt, tot;
struct unq {
  int val, id, op;
} u[4 * MAXN];
struct tr {
  int cov, l, r, sum;
} t[16 * MAXN], an;
int ans;
bool cmp(const unq &a, const unq &b) { return a.val < b.val; }
tr operator+(const tr &a, const tr &b) {
  if (a.r == b.l && a.r) {
    return (tr){0, a.l, b.r, a.sum + b.sum - 2};
  }
  return (tr){0, a.l, b.r, a.sum + b.sum};
}
void update(int l, int r, int z, int y, int v, int id) {
  if (l == z && r == y) {
    t[id].cov += v;
    if (!t[id].cov) {
      if (l == r) {
        t[id].l = t[id].r = 0;
        t[id].sum = 0;
      } else
        t[id] = t[id * 2] + t[id * 2 + 1];
    } else
      t[id].r = t[id].l = 1, t[id].sum = 2;
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= y)
    update(l, mid, z, y, v, id * 2);
  else if (mid < z)
    update(mid + 1, r, z, y, v, id * 2 + 1);
  else {
    update(l, mid, z, mid, v, id * 2);
    update(mid + 1, r, mid + 1, y, v, id * 2 + 1);
  }
  if (!t[id].cov) t[id] = t[id * 2] + t[id * 2 + 1];
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    ans = cnt = tot = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &s[i], &e[i]);
      s[i] *= 2;
      e[i] *= 2;
      cnt++;
      u[cnt].val = s[i];
      u[cnt].id = i;
      u[cnt].op = 1;
      cnt++;
      u[cnt].val = s[i] - 1;
      u[cnt].id = i;
      u[cnt].op = 3;
      cnt++;
      u[cnt].val = e[i];
      u[cnt].id = i;
      u[cnt].op = 2;
      cnt++;
      u[cnt].val = e[i] + 1;
      u[cnt].id = i;
      u[cnt].op = 3;
    }
    sort(u + 1, u + 1 + cnt, cmp);
    for (int i = 1; i <= cnt; i++) {
      if (i == 1 || u[i].val != u[i - 1].val) tot++;
      if (u[i].op == 1) s[u[i].id] = tot;
      if (u[i].op == 2) e[u[i].id] = tot;
    }
    for (int i = 1; i <= n; i++) {
      update(1, tot, s[i], e[i], 1, 1);
    }
    for (int i = 1; i <= n; i++) {
      update(1, tot, s[i], e[i], -1, 1);
      ans = max(ans, t[1].sum - 2);
      update(1, tot, s[i], e[i], 1, 1);
    }
    printf("%d\n", ans / 2 + 1);
    for (int i = 1; i <= n; i++) update(1, tot, s[i], e[i], -1, 1);
  }
  return 0;
}

```