#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
using VI = vector<int>;
struct SegmentTree {
  int sz;
  vector<int> low, lowcnt, lazy;
  vector<int> lt, rt;
  void split(int node) {
    low[2 * node] += lazy[node];
    lazy[2 * node] += lazy[node];
    lt[2 * node] += lazy[node];
    rt[2 * node] += lazy[node];
    low[2 * node + 1] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
    lt[2 * node + 1] += lazy[node];
    rt[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
  void merge(int node) {
    if (low[2 * node] == low[2 * node + 1]) {
      low[node] = low[2 * node];
      lowcnt[node] = lowcnt[2 * node] + lowcnt[2 * node + 1];
      if (rt[2 * node] == lt[2 * node + 1] && rt[2 * node] == low[node]) {
        lowcnt[node]--;
      }
    } else {
      int ch = (low[2 * node] < low[2 * node + 1]) ? 2 * node : 2 * node + 1;
      low[node] = low[ch];
      lowcnt[node] = lowcnt[ch];
    }
    lt[node] = lt[2 * node];
    rt[node] = rt[2 * node + 1];
  }
  SegmentTree(int n) {
    for (sz = 1; sz < n; sz <<= 1)
      ;
    low.assign(2 * sz, 0);
    lowcnt.assign(2 * sz, 1);
    lazy.assign(2 * sz, 0);
    lt.assign(2 * sz, 0);
    rt.assign(2 * sz, 0);
    for (int i = sz - 1; i >= 1; i--) {
      merge(i);
    }
  }
  void update(int l, int r, int d, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (r <= x || l >= y) return;
    if (l <= x && y <= r) {
      low[node] += d;
      lazy[node] += d;
      lt[node] += d;
      rt[node] += d;
      return;
    }
    split(node);
    update(l, r, d, 2 * node, x, (x + y) / 2);
    update(l, r, d, 2 * node + 1, (x + y) / 2, y);
    merge(node);
  }
  int get() {
    assert(low[1] == 0);
    return lowcnt[1];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> pts;
    vector<PII> ranges;
    for (int i = 0; i < n; i++) {
      int l, r;
      cin >> l >> r;
      pts.push_back(l);
      pts.push_back(r);
      ranges.emplace_back(l, r);
    }
    sort(pts.begin(), pts.end());
    pts.resize(unique(pts.begin(), pts.end()) - pts.begin());
    SegmentTree st(2 * pts.size() + 4);
    for (auto& e : ranges) {
      int l = lower_bound(pts.begin(), pts.end(), e.first) - pts.begin();
      int r = lower_bound(pts.begin(), pts.end(), e.second) - pts.begin();
      l = 2 * l + 2;
      r = 2 * r + 2;
      st.update(l, r + 1, +1);
      e = make_pair(l, r);
    }
    int ans = 0;
    for (auto& e : ranges) {
      st.update(e.first, e.second + 1, -1);
      ans = max(ans, st.get() - 1);
      st.update(e.first, e.second + 1, +1);
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
using VI = vector<int>;
struct SegmentTree {
  int sz;
  vector<int> low, lowcnt, lazy;
  vector<int> lt, rt;
  void split(int node) {
    low[2 * node] += lazy[node];
    lazy[2 * node] += lazy[node];
    lt[2 * node] += lazy[node];
    rt[2 * node] += lazy[node];
    low[2 * node + 1] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
    lt[2 * node + 1] += lazy[node];
    rt[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
  void merge(int node) {
    if (low[2 * node] == low[2 * node + 1]) {
      low[node] = low[2 * node];
      lowcnt[node] = lowcnt[2 * node] + lowcnt[2 * node + 1];
      if (rt[2 * node] == lt[2 * node + 1] && rt[2 * node] == low[node]) {
        lowcnt[node]--;
      }
    } else {
      int ch = (low[2 * node] < low[2 * node + 1]) ? 2 * node : 2 * node + 1;
      low[node] = low[ch];
      lowcnt[node] = lowcnt[ch];
    }
    lt[node] = lt[2 * node];
    rt[node] = rt[2 * node + 1];
  }
  SegmentTree(int n) {
    for (sz = 1; sz < n; sz <<= 1)
      ;
    low.assign(2 * sz, 0);
    lowcnt.assign(2 * sz, 1);
    lazy.assign(2 * sz, 0);
    lt.assign(2 * sz, 0);
    rt.assign(2 * sz, 0);
    for (int i = sz - 1; i >= 1; i--) {
      merge(i);
    }
  }
  void update(int l, int r, int d, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (r <= x || l >= y) return;
    if (l <= x && y <= r) {
      low[node] += d;
      lazy[node] += d;
      lt[node] += d;
      rt[node] += d;
      return;
    }
    split(node);
    update(l, r, d, 2 * node, x, (x + y) / 2);
    update(l, r, d, 2 * node + 1, (x + y) / 2, y);
    merge(node);
  }
  int get() {
    assert(low[1] == 0);
    return lowcnt[1];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> pts;
    vector<PII> ranges;
    for (int i = 0; i < n; i++) {
      int l, r;
      cin >> l >> r;
      pts.push_back(l);
      pts.push_back(r);
      ranges.emplace_back(l, r);
    }
    sort(pts.begin(), pts.end());
    pts.resize(unique(pts.begin(), pts.end()) - pts.begin());
    SegmentTree st(2 * pts.size() + 4);
    for (auto& e : ranges) {
      int l = lower_bound(pts.begin(), pts.end(), e.first) - pts.begin();
      int r = lower_bound(pts.begin(), pts.end(), e.second) - pts.begin();
      l = 2 * l + 2;
      r = 2 * r + 2;
      st.update(l, r + 1, +1);
      e = make_pair(l, r);
    }
    int ans = 0;
    for (auto& e : ranges) {
      st.update(e.first, e.second + 1, -1);
      ans = max(ans, st.get() - 1);
      st.update(e.first, e.second + 1, +1);
    }
    cout << ans << '\n';
  }
  return 0;
}

```