#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct Universe {
  vector<T> o;
  void push(const T &x) { o.push_back(x); }
  int build() {
    sort(o.begin(), o.end());
    o.resize(unique(o.begin(), o.end()) - o.begin());
    return o.size();
  }
  int find(const T &x) {
    auto it = lower_bound(o.begin(), o.end(), x);
    return (it == o.end() || x < *it ? -1 : it - o.begin() + 1);
  }
  T &operator[](int i) { return o[i - 1]; }
};
struct Node {
  int freq;
  bool left, right;
  Node(bool b) { freq = left = right = b; }
  Node operator+(const Node &o) {
    Node r(0);
    r.freq = freq + o.freq - (right & o.left);
    r.left = left;
    r.right = o.right;
    return r;
  }
} NIL(0);
template <class T>
struct SegTree {
  vector<Node> tree;
  SegTree(int n) : tree(n << 2, NIL) {}
  void update(int i, int l, int r, int p, T x) {
    if (p > r || p < l) {
      return;
    } else if (l == r) {
      tree[i] = Node(x);
    } else {
      int m = (l + r) >> 1;
      int fe = i << 1, fd = (i << 1) + 1;
      update(fe, l, m, p, x);
      update(fd, m + 1, r, p, x);
      tree[i] = tree[fe] + tree[fd];
    }
  }
  Node query(int i, int l, int r, int ql, int qr) {
    if (r < ql || l > qr) {
      return NIL;
    } else if (l >= ql && r <= qr) {
      return tree[i];
    } else {
      int m = (l + r) >> 1;
      int fe = i << 1, fd = (i << 1) + 1;
      return query(fe, l, m, ql, qr) + query(fd, m + 1, r, ql, qr);
    }
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<pair<int, int> > in;
    Universe<int> cc;
    for (int i = 0; i < n; i++) {
      int l, r;
      scanf("%d %d", &l, &r);
      cc.push(l);
      cc.push(r);
      in.push_back({l, r});
    }
    int m = (cc.build() << 1) + 5;
    vector<int> open(m), close(m);
    for (int i = 0; i < n; i++) {
      open[cc.find(in[i].first) << 1]++;
      close[cc.find(in[i].second) << 1]++;
    }
    vector<int> freq(m);
    for (int i = 0, total = 0; i < m; i++) {
      total += open[i];
      freq[i] = total;
      total -= close[i];
    }
    vector<int> left(m);
    for (int i = 1; i < m; i++) {
      if (freq[i] == 1) {
        left[i] = left[i - 1];
      } else {
        left[i] = freq[i];
      }
    }
    vector<int> right(m);
    for (int i = m - 2; i >= 0; i--) {
      if (freq[i] == 1) {
        right[i] = right[i + 1];
      } else {
        right[i] = freq[i];
      }
    }
    vector<bool> loose(m);
    int all = 0;
    SegTree<bool> t(m);
    for (int i = 1; i < m; i++) {
      if (freq[i] == 1) {
        if (left[i] == 0 && right[i] == 0) {
          loose[i] = true;
        } else if (left[i] > 0 && right[i] > 0) {
          t.update(1, 0, m - 1, i, true);
        }
      } else if (freq[i] == 0 && freq[i - 1] > 0) {
        all++;
      }
    }
    int global = 0;
    for (int i = 0; i < n; i++) {
      int ql = cc.find(in[i].first) << 1;
      int qr = cc.find(in[i].second) << 1;
      int add = (loose[ql] ? -1 : t.query(1, 0, m - 1, ql, qr).freq);
      global = max(global, all + add);
    }
    printf("%d\n", global);
  }
}

### 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;
template <typename T>
struct Universe {
  vector<T> o;
  void push(const T &x) { o.push_back(x); }
  int build() {
    sort(o.begin(), o.end());
    o.resize(unique(o.begin(), o.end()) - o.begin());
    return o.size();
  }
  int find(const T &x) {
    auto it = lower_bound(o.begin(), o.end(), x);
    return (it == o.end() || x < *it ? -1 : it - o.begin() + 1);
  }
  T &operator[](int i) { return o[i - 1]; }
};
struct Node {
  int freq;
  bool left, right;
  Node(bool b) { freq = left = right = b; }
  Node operator+(const Node &o) {
    Node r(0);
    r.freq = freq + o.freq - (right & o.left);
    r.left = left;
    r.right = o.right;
    return r;
  }
} NIL(0);
template <class T>
struct SegTree {
  vector<Node> tree;
  SegTree(int n) : tree(n << 2, NIL) {}
  void update(int i, int l, int r, int p, T x) {
    if (p > r || p < l) {
      return;
    } else if (l == r) {
      tree[i] = Node(x);
    } else {
      int m = (l + r) >> 1;
      int fe = i << 1, fd = (i << 1) + 1;
      update(fe, l, m, p, x);
      update(fd, m + 1, r, p, x);
      tree[i] = tree[fe] + tree[fd];
    }
  }
  Node query(int i, int l, int r, int ql, int qr) {
    if (r < ql || l > qr) {
      return NIL;
    } else if (l >= ql && r <= qr) {
      return tree[i];
    } else {
      int m = (l + r) >> 1;
      int fe = i << 1, fd = (i << 1) + 1;
      return query(fe, l, m, ql, qr) + query(fd, m + 1, r, ql, qr);
    }
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<pair<int, int> > in;
    Universe<int> cc;
    for (int i = 0; i < n; i++) {
      int l, r;
      scanf("%d %d", &l, &r);
      cc.push(l);
      cc.push(r);
      in.push_back({l, r});
    }
    int m = (cc.build() << 1) + 5;
    vector<int> open(m), close(m);
    for (int i = 0; i < n; i++) {
      open[cc.find(in[i].first) << 1]++;
      close[cc.find(in[i].second) << 1]++;
    }
    vector<int> freq(m);
    for (int i = 0, total = 0; i < m; i++) {
      total += open[i];
      freq[i] = total;
      total -= close[i];
    }
    vector<int> left(m);
    for (int i = 1; i < m; i++) {
      if (freq[i] == 1) {
        left[i] = left[i - 1];
      } else {
        left[i] = freq[i];
      }
    }
    vector<int> right(m);
    for (int i = m - 2; i >= 0; i--) {
      if (freq[i] == 1) {
        right[i] = right[i + 1];
      } else {
        right[i] = freq[i];
      }
    }
    vector<bool> loose(m);
    int all = 0;
    SegTree<bool> t(m);
    for (int i = 1; i < m; i++) {
      if (freq[i] == 1) {
        if (left[i] == 0 && right[i] == 0) {
          loose[i] = true;
        } else if (left[i] > 0 && right[i] > 0) {
          t.update(1, 0, m - 1, i, true);
        }
      } else if (freq[i] == 0 && freq[i - 1] > 0) {
        all++;
      }
    }
    int global = 0;
    for (int i = 0; i < n; i++) {
      int ql = cc.find(in[i].first) << 1;
      int qr = cc.find(in[i].second) << 1;
      int add = (loose[ql] ? -1 : t.query(1, 0, m - 1, ql, qr).freq);
      global = max(global, all + add);
    }
    printf("%d\n", global);
  }
}

```