#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int man = 2e5 + 10;
template <typename T>
inline T read() {
  T sum = 0, fl = 1;
  int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fl = -1;
  for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
  return sum * fl;
}
template <typename T>
inline void write(T x) {
  static int sta[35];
  int top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(sta[--top] + 48);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T exgcd(T a, T b, T &g, T &x, T &y) {
  if (!b) {
    g = a, x = 1, y = 0;
  } else {
    exgcd(b, a % b, g, y, x);
    y -= x * (a / b);
  }
}
const long long mod = 1e9 + 7;
pair<int, int> sp[man];
int num[man], sta[man], max_r[man];
int top;
int bin(int x) {
  int l = 1, r = top;
  int ans = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (sta[mid] >= x) {
      l = mid + 1;
      ans = mid;
    } else
      r = mid - 1;
  }
  if (sta[ans] == x) ans--;
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    max_r[0] = -INT_MAX;
    num[0] = 0;
    top = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &sp[i].first, &sp[i].second);
    }
    sort(sp + 1, sp + 1 + n);
    for (int i = 1; i <= n; ++i) {
      max_r[i] = max(sp[i].second, max_r[i - 1]);
      if (sp[i].first > max_r[i - 1]) num[i] = 1;
      num[i] += num[i - 1];
    }
    int ans = 0;
    for (int i = n; i >= 1; i--) {
      ans = max(ans, num[i - 1] + bin(max_r[i - 1]));
      while (top && sp[i].second >= sta[top]) top--;
      sta[++top] = sp[i].first;
      num[i] = 0;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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>
#pragma GCC optimize(2)
using namespace std;
const int man = 2e5 + 10;
template <typename T>
inline T read() {
  T sum = 0, fl = 1;
  int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fl = -1;
  for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
  return sum * fl;
}
template <typename T>
inline void write(T x) {
  static int sta[35];
  int top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(sta[--top] + 48);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T exgcd(T a, T b, T &g, T &x, T &y) {
  if (!b) {
    g = a, x = 1, y = 0;
  } else {
    exgcd(b, a % b, g, y, x);
    y -= x * (a / b);
  }
}
const long long mod = 1e9 + 7;
pair<int, int> sp[man];
int num[man], sta[man], max_r[man];
int top;
int bin(int x) {
  int l = 1, r = top;
  int ans = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (sta[mid] >= x) {
      l = mid + 1;
      ans = mid;
    } else
      r = mid - 1;
  }
  if (sta[ans] == x) ans--;
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    max_r[0] = -INT_MAX;
    num[0] = 0;
    top = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &sp[i].first, &sp[i].second);
    }
    sort(sp + 1, sp + 1 + n);
    for (int i = 1; i <= n; ++i) {
      max_r[i] = max(sp[i].second, max_r[i - 1]);
      if (sp[i].first > max_r[i - 1]) num[i] = 1;
      num[i] += num[i - 1];
    }
    int ans = 0;
    for (int i = n; i >= 1; i--) {
      ans = max(ans, num[i - 1] + bin(max_r[i - 1]));
      while (top && sp[i].second >= sta[top]) top--;
      sta[++top] = sp[i].first;
      num[i] = 0;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```