#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int N = 3e6 + 100;
pair<int, pair<bool, bool>> t[N << 2];
const long long oo = 1e18;
int a[N];
pair<int, pair<bool, bool>> trans(pair<int, pair<bool, bool>> a,
                                  pair<int, pair<bool, bool>> b) {
  int cnt = a.first + b.first;
  if (b.second.first && a.second.second) cnt--;
  return {cnt, {a.second.first, b.second.second}};
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    if (a[tl] == 1) {
      t[v] = {1, {1, 1}};
    } else {
      t[v] = {0, {0, 0}};
    }
    return;
  }
  int tm = tl + tr >> 1;
  build(v << 1, tl, tm);
  build(v << 1 | 1, tm + 1, tr);
  t[v] = trans(t[v << 1], t[v << 1 | 1]);
}
pair<int, pair<bool, bool>> get(int v, int tl, int tr, int l, int r) {
  assert(l <= r);
  if (tl == l && tr == r) return t[v];
  int tm = tl + tr >> 1;
  if (l <= tm && r > tm)
    return trans(get(v << 1, tl, tm, l, min(tm, r)),
                 get(v << 1 | 1, tm + 1, tr, max(tm + 1, l), r));
  if (l <= tm) return get(v << 1, tl, tm, l, min(tm, r));
  return get(v << 1 | 1, tm + 1, tr, max(tm + 1, l), r);
}
int p[N];
void solve() {
  int n;
  cin >> n;
  vector<pair<long long, long long>> segs(n);
  vector<long long> points;
  for (int i = 0; i < int(n); ++i) {
    cin >> segs[i].first >> segs[i].second;
    segs[i].first *= 2, segs[i].second *= 2;
    for (int dif = -1; dif <= 1; ++dif) {
      points.push_back(segs[i].first + dif);
      points.push_back(segs[i].second + dif);
    }
  }
  points.push_back(-oo);
  sort(points.begin(), points.end());
  points.resize(unique(points.begin(), points.end()) - points.begin());
  int NN = 0;
  for (int i = 0; i < int(n); ++i) {
    segs[i].first = lower_bound(points.begin(), points.end(), segs[i].first) -
                    points.begin();
    segs[i].second = lower_bound(points.begin(), points.end(), segs[i].second) -
                     points.begin();
    NN = max(NN, ((int)segs[i].second + 10));
  }
  for (int i = 0; i < int(NN); ++i) a[i] = 0;
  for (int i = 0; i < int(n); ++i) {
    a[segs[i].first]++;
    a[segs[i].second + 1]--;
  }
  int all = 0;
  p[0] = a[0];
  for (int i = 1; i < NN; ++i) {
    a[i] += a[i - 1];
    p[i] = a[i] + p[i - 1];
    if (a[i - 1] == 0 && a[i]) all++;
  }
  build(1, 0, NN - 1);
  int ans = 0;
  for (int i = 0; i < int(n); ++i) {
    pair<int, pair<bool, bool>> res =
        get(1, 0, NN - 1, segs[i].first, segs[i].second);
    int cur = res.first;
    if (res.second.first && a[segs[i].first - 1] == 0) {
      cur--;
    }
    if (res.second.second && a[segs[i].second + 1] == 0) {
      cur--;
    }
    if (p[segs[i].second] - p[segs[i].first - 1] ==
        segs[i].second - segs[i].first + 1)
      cur = -1;
    ans = max(ans, (all + cur));
  }
  cout << ans << '\n';
}
void smain() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

In cpp, your task is to solve 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;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int N = 3e6 + 100;
pair<int, pair<bool, bool>> t[N << 2];
const long long oo = 1e18;
int a[N];
pair<int, pair<bool, bool>> trans(pair<int, pair<bool, bool>> a,
                                  pair<int, pair<bool, bool>> b) {
  int cnt = a.first + b.first;
  if (b.second.first && a.second.second) cnt--;
  return {cnt, {a.second.first, b.second.second}};
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    if (a[tl] == 1) {
      t[v] = {1, {1, 1}};
    } else {
      t[v] = {0, {0, 0}};
    }
    return;
  }
  int tm = tl + tr >> 1;
  build(v << 1, tl, tm);
  build(v << 1 | 1, tm + 1, tr);
  t[v] = trans(t[v << 1], t[v << 1 | 1]);
}
pair<int, pair<bool, bool>> get(int v, int tl, int tr, int l, int r) {
  assert(l <= r);
  if (tl == l && tr == r) return t[v];
  int tm = tl + tr >> 1;
  if (l <= tm && r > tm)
    return trans(get(v << 1, tl, tm, l, min(tm, r)),
                 get(v << 1 | 1, tm + 1, tr, max(tm + 1, l), r));
  if (l <= tm) return get(v << 1, tl, tm, l, min(tm, r));
  return get(v << 1 | 1, tm + 1, tr, max(tm + 1, l), r);
}
int p[N];
void solve() {
  int n;
  cin >> n;
  vector<pair<long long, long long>> segs(n);
  vector<long long> points;
  for (int i = 0; i < int(n); ++i) {
    cin >> segs[i].first >> segs[i].second;
    segs[i].first *= 2, segs[i].second *= 2;
    for (int dif = -1; dif <= 1; ++dif) {
      points.push_back(segs[i].first + dif);
      points.push_back(segs[i].second + dif);
    }
  }
  points.push_back(-oo);
  sort(points.begin(), points.end());
  points.resize(unique(points.begin(), points.end()) - points.begin());
  int NN = 0;
  for (int i = 0; i < int(n); ++i) {
    segs[i].first = lower_bound(points.begin(), points.end(), segs[i].first) -
                    points.begin();
    segs[i].second = lower_bound(points.begin(), points.end(), segs[i].second) -
                     points.begin();
    NN = max(NN, ((int)segs[i].second + 10));
  }
  for (int i = 0; i < int(NN); ++i) a[i] = 0;
  for (int i = 0; i < int(n); ++i) {
    a[segs[i].first]++;
    a[segs[i].second + 1]--;
  }
  int all = 0;
  p[0] = a[0];
  for (int i = 1; i < NN; ++i) {
    a[i] += a[i - 1];
    p[i] = a[i] + p[i - 1];
    if (a[i - 1] == 0 && a[i]) all++;
  }
  build(1, 0, NN - 1);
  int ans = 0;
  for (int i = 0; i < int(n); ++i) {
    pair<int, pair<bool, bool>> res =
        get(1, 0, NN - 1, segs[i].first, segs[i].second);
    int cur = res.first;
    if (res.second.first && a[segs[i].first - 1] == 0) {
      cur--;
    }
    if (res.second.second && a[segs[i].second + 1] == 0) {
      cur--;
    }
    if (p[segs[i].second] - p[segs[i].first - 1] ==
        segs[i].second - segs[i].first + 1)
      cur = -1;
    ans = max(ans, (all + cur));
  }
  cout << ans << '\n';
}
void smain() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```