#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  register long long x = 0, f = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10ll + (ch ^ 48), ch = getchar();
  return x * f;
}
void Write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) Write(x / 10);
  putchar(x % 10 + 48);
}
void writes(long long x) { Write(x), putchar(' '); }
void Writes(long long x) { Write(x), putchar('\n'); }
int n;
struct node {
  int l, r;
} Seg[201010];
int Map[201010 << 1], tot, Max;
int sum[201010 << 2], from[201010 << 2], lf[201010], rf[201010];
int tans, ans;
bool two = 0;
void Reset() {
  for (int i = 1; i <= Max; ++i) sum[i] = 0, from[i] = 0;
  for (int i = 1; i <= Max; ++i) lf[i] = rf[i] = 0;
  two = 0, tot = 0, Max = 0, tans = 0, ans = 0;
}
int Find(int x) { return lower_bound(Map + 1, Map + 1 + tot, x) - Map; }
void Treat() {
  sort(Map + 1, Map + 1 + tot), tot = unique(Map + 1, Map + 1 + tot) - Map - 1;
  for (int i = 1; i <= n; ++i) {
    Seg[i].l = Find(Seg[i].l) << 1, Seg[i].r = Find(Seg[i].r) << 1;
    sum[Seg[i].l]++, sum[Seg[i].r + 1]--;
    from[Seg[i].l] += i, from[Seg[i].r + 1] -= i;
    Max = max(Max, Seg[i].r + 1);
  }
}
int GetAns1() {
  bool state = 0;
  int tmp = 0;
  for (int i = 1; i <= Max; ++i) {
    sum[i] += sum[i - 1], from[i] += from[i - 1];
    if (sum[i] > 1) two = 1;
  }
  for (int i = 1; i <= Max; ++i) {
    if (sum[i] == 1) {
      if (!lf[from[i]])
        lf[from[i]] = rf[from[i]] = i;
      else
        rf[from[i]] = i;
    }
  }
  for (int i = 1; i <= Max; ++i) {
    if (sum[i] != 0) {
      if (state == 0)
        tmp++, state ^= 1;
      else
        ;
    } else if (sum[i] == 0)
      if (state == 1) state ^= 1;
  }
  return tmp;
}
void GetAns2() {
  int add = 0, state = 0, id;
  for (int i = 1; i <= Max;) {
    if (sum[i] == 1) {
      id = from[i];
      int l = lf[id], r = rf[id];
      state = 0, add = 0;
      for (int loc = l; loc <= r; ++loc) {
        if (sum[loc] > 1) {
          if (state == 0)
            add++, state ^= 1;
          else
            ;
        } else if (sum[loc] == 1)
          if (state == 1) state ^= 1;
      }
      if (sum[l - 1] > 1 and sum[r + 1] > 1)
        add++;
      else if (sum[l - 1] > 1 or sum[r + 1] > 1)
        ;
      else if (sum[l - 1] <= 1 and sum[r + 1] <= 1)
        add--;
      ans = max(ans, tans + add);
      i = r + 1;
    } else
      ++i;
  }
}
int main() {
  int T = read();
  while (T--) {
    Reset();
    n = read();
    for (int i = 1; i <= n; ++i)
      Map[++tot] = Seg[i].l = read(), Map[++tot] = Seg[i].r = read();
    Treat();
    tans = GetAns1();
    if (!two)
      ans = tans - 1;
    else
      ans = tans;
    GetAns2();
    Writes(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;
inline long long read() {
  register long long x = 0, f = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10ll + (ch ^ 48), ch = getchar();
  return x * f;
}
void Write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) Write(x / 10);
  putchar(x % 10 + 48);
}
void writes(long long x) { Write(x), putchar(' '); }
void Writes(long long x) { Write(x), putchar('\n'); }
int n;
struct node {
  int l, r;
} Seg[201010];
int Map[201010 << 1], tot, Max;
int sum[201010 << 2], from[201010 << 2], lf[201010], rf[201010];
int tans, ans;
bool two = 0;
void Reset() {
  for (int i = 1; i <= Max; ++i) sum[i] = 0, from[i] = 0;
  for (int i = 1; i <= Max; ++i) lf[i] = rf[i] = 0;
  two = 0, tot = 0, Max = 0, tans = 0, ans = 0;
}
int Find(int x) { return lower_bound(Map + 1, Map + 1 + tot, x) - Map; }
void Treat() {
  sort(Map + 1, Map + 1 + tot), tot = unique(Map + 1, Map + 1 + tot) - Map - 1;
  for (int i = 1; i <= n; ++i) {
    Seg[i].l = Find(Seg[i].l) << 1, Seg[i].r = Find(Seg[i].r) << 1;
    sum[Seg[i].l]++, sum[Seg[i].r + 1]--;
    from[Seg[i].l] += i, from[Seg[i].r + 1] -= i;
    Max = max(Max, Seg[i].r + 1);
  }
}
int GetAns1() {
  bool state = 0;
  int tmp = 0;
  for (int i = 1; i <= Max; ++i) {
    sum[i] += sum[i - 1], from[i] += from[i - 1];
    if (sum[i] > 1) two = 1;
  }
  for (int i = 1; i <= Max; ++i) {
    if (sum[i] == 1) {
      if (!lf[from[i]])
        lf[from[i]] = rf[from[i]] = i;
      else
        rf[from[i]] = i;
    }
  }
  for (int i = 1; i <= Max; ++i) {
    if (sum[i] != 0) {
      if (state == 0)
        tmp++, state ^= 1;
      else
        ;
    } else if (sum[i] == 0)
      if (state == 1) state ^= 1;
  }
  return tmp;
}
void GetAns2() {
  int add = 0, state = 0, id;
  for (int i = 1; i <= Max;) {
    if (sum[i] == 1) {
      id = from[i];
      int l = lf[id], r = rf[id];
      state = 0, add = 0;
      for (int loc = l; loc <= r; ++loc) {
        if (sum[loc] > 1) {
          if (state == 0)
            add++, state ^= 1;
          else
            ;
        } else if (sum[loc] == 1)
          if (state == 1) state ^= 1;
      }
      if (sum[l - 1] > 1 and sum[r + 1] > 1)
        add++;
      else if (sum[l - 1] > 1 or sum[r + 1] > 1)
        ;
      else if (sum[l - 1] <= 1 and sum[r + 1] <= 1)
        add--;
      ans = max(ans, tans + add);
      i = r + 1;
    } else
      ++i;
  }
}
int main() {
  int T = read();
  while (T--) {
    Reset();
    n = read();
    for (int i = 1; i <= n; ++i)
      Map[++tot] = Seg[i].l = read(), Map[++tot] = Seg[i].r = read();
    Treat();
    tans = GetAns1();
    if (!two)
      ans = tans - 1;
    else
      ans = tans;
    GetAns2();
    Writes(ans);
  }
  return 0;
}

```