#include <bits/stdc++.h>
#pragma GCC optimize("O3,Ofast,fast-math,unroll-loops,no-stack-protector")
using namespace std;
long long sz;
vector<long long> sum;
vector<long long> pref;
long long next(long long n) {
  long long i = 1;
  while (i < n) {
    i <<= 1;
  }
  return i;
}
struct Node {
  long long sz, cnt;
  bool hl, hr;
  Node() {}
  Node(long long _sz, bool _hl, bool _hr, long long _cnt) {
    sz = _sz;
    hl = _hl;
    hr = _hr;
    cnt = _cnt;
  }
};
Node join(Node a, Node b) {
  if (a.sz == -1) {
    return b;
  } else if (b.sz == -1) {
    return a;
  }
  long long delta = 0;
  if (a.hr && b.hl) {
    delta = -1;
  }
  return Node(a.sz + b.sz, a.hl, b.hr, a.cnt + b.cnt + delta);
}
vector<Node> tree;
Node get(long long i, long long L, long long R, long long l, long long r) {
  if (l <= L && R <= r) {
    return tree[i];
  } else if (!(l > R || L > r)) {
    long long M = (L + R) >> 1;
    Node _l = get(i << 1, L, M, l, r);
    Node _r = get(i << 1 | 1, M + 1, R, l, r);
    Node _j = join(_l, _r);
    return _j;
  } else {
    return Node(-1, false, false, 0);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t-- > 0) {
    long long n;
    cin >> n;
    sz = next((n << 2) + 1);
    sum.resize(sz);
    pref.resize(sz);
    tree.resize(sz << 1);
    vector<pair<long long, long long>> kek(n);
    set<long long> used;
    map<long long, long long> to;
    for (long long i = 0; i < n; i++) {
      long long l, r;
      cin >> l >> r;
      used.insert(l);
      used.insert(r);
      kek[i] = {l, r};
    }
    long long now = 2;
    for (long long x : used) {
      to[x] = now;
      now += 2;
    }
    for (long long i = 0; i < sz; i++) {
      sum[i] = 0;
      pref[i] = 0;
    }
    for (long long i = 0; i < n; i++) {
      kek[i].first = to[kek[i].first];
      kek[i].second = to[kek[i].second];
      pref[kek[i].first]++;
      pref[kek[i].second + 1]--;
    }
    long long curr = 0;
    for (long long i = 0; i < sz; i++) {
      curr += pref[i];
      sum[i] = curr;
    }
    long long ans = 1;
    sort(kek.begin(), kek.end());
    long long xl = kek[0].first;
    long long xr = kek[0].second;
    for (long long i = 1; i < kek.size(); i++) {
      if (kek[i].first > xr) {
        ans++;
      }
      xr = max(xr, kek[i].second);
    }
    for (long long i = 0; i < sz; i++) {
      if (sum[i] == 1) {
        tree[sz + i] = {1, true, true, 1};
      } else {
        tree[sz + i] = {1, false, false, 0};
      }
    }
    for (long long i = sz - 1; i >= 1; i--) {
      tree[i] = join(tree[i << 1], tree[i << 1 | 1]);
    }
    long long best = 0;
    for (pair<long long, long long> pr : kek) {
      Node now = get(1, 0, sz - 1, pr.first, pr.second);
      if (now.cnt == 1 && sum[pr.first] == 1 && sum[pr.second] == 1) {
        best = max(best, ans - 1);
        continue;
      } else {
        long long poss = ans + now.cnt;
        if (sum[pr.first] == 1 && now.hl) {
          poss--;
        }
        if (sum[pr.second] == 1 && now.hr) {
          poss--;
        }
        best = max(best, poss);
      }
    }
    cout << best << '\n';
  }
  fflush(stdout);
}

### Prompt

Your task is to create 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("O3,Ofast,fast-math,unroll-loops,no-stack-protector")
using namespace std;
long long sz;
vector<long long> sum;
vector<long long> pref;
long long next(long long n) {
  long long i = 1;
  while (i < n) {
    i <<= 1;
  }
  return i;
}
struct Node {
  long long sz, cnt;
  bool hl, hr;
  Node() {}
  Node(long long _sz, bool _hl, bool _hr, long long _cnt) {
    sz = _sz;
    hl = _hl;
    hr = _hr;
    cnt = _cnt;
  }
};
Node join(Node a, Node b) {
  if (a.sz == -1) {
    return b;
  } else if (b.sz == -1) {
    return a;
  }
  long long delta = 0;
  if (a.hr && b.hl) {
    delta = -1;
  }
  return Node(a.sz + b.sz, a.hl, b.hr, a.cnt + b.cnt + delta);
}
vector<Node> tree;
Node get(long long i, long long L, long long R, long long l, long long r) {
  if (l <= L && R <= r) {
    return tree[i];
  } else if (!(l > R || L > r)) {
    long long M = (L + R) >> 1;
    Node _l = get(i << 1, L, M, l, r);
    Node _r = get(i << 1 | 1, M + 1, R, l, r);
    Node _j = join(_l, _r);
    return _j;
  } else {
    return Node(-1, false, false, 0);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t-- > 0) {
    long long n;
    cin >> n;
    sz = next((n << 2) + 1);
    sum.resize(sz);
    pref.resize(sz);
    tree.resize(sz << 1);
    vector<pair<long long, long long>> kek(n);
    set<long long> used;
    map<long long, long long> to;
    for (long long i = 0; i < n; i++) {
      long long l, r;
      cin >> l >> r;
      used.insert(l);
      used.insert(r);
      kek[i] = {l, r};
    }
    long long now = 2;
    for (long long x : used) {
      to[x] = now;
      now += 2;
    }
    for (long long i = 0; i < sz; i++) {
      sum[i] = 0;
      pref[i] = 0;
    }
    for (long long i = 0; i < n; i++) {
      kek[i].first = to[kek[i].first];
      kek[i].second = to[kek[i].second];
      pref[kek[i].first]++;
      pref[kek[i].second + 1]--;
    }
    long long curr = 0;
    for (long long i = 0; i < sz; i++) {
      curr += pref[i];
      sum[i] = curr;
    }
    long long ans = 1;
    sort(kek.begin(), kek.end());
    long long xl = kek[0].first;
    long long xr = kek[0].second;
    for (long long i = 1; i < kek.size(); i++) {
      if (kek[i].first > xr) {
        ans++;
      }
      xr = max(xr, kek[i].second);
    }
    for (long long i = 0; i < sz; i++) {
      if (sum[i] == 1) {
        tree[sz + i] = {1, true, true, 1};
      } else {
        tree[sz + i] = {1, false, false, 0};
      }
    }
    for (long long i = sz - 1; i >= 1; i--) {
      tree[i] = join(tree[i << 1], tree[i << 1 | 1]);
    }
    long long best = 0;
    for (pair<long long, long long> pr : kek) {
      Node now = get(1, 0, sz - 1, pr.first, pr.second);
      if (now.cnt == 1 && sum[pr.first] == 1 && sum[pr.second] == 1) {
        best = max(best, ans - 1);
        continue;
      } else {
        long long poss = ans + now.cnt;
        if (sum[pr.first] == 1 && now.hl) {
          poss--;
        }
        if (sum[pr.second] == 1 && now.hr) {
          poss--;
        }
        best = max(best, poss);
      }
    }
    cout << best << '\n';
  }
  fflush(stdout);
}

```