#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 6;
int T;
int n, tot, x, cnt;
map<int, int> mapp;
vector<int> v;
template <class T>
void read(T &x, int f = 1) {
  char ch = getchar();
  x = 0;
  while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : 1), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
struct node {
  node *lson, *rson;
  int cnt, cover, l, r;
  int mxl, mxr;
} tr[maxn * 8], *root;
struct segment {
  int l, r;
} s[maxn];
inline node *Newnode(node *&root) { return root = &tr[tot++]; }
inline void update(node *cur) {
  if (cur->cover)
    cur->cnt = 1, cur->mxl = cur->l, cur->mxr = cur->r;
  else {
    cur->mxl = cur->mxr = 0;
    if (cur->l == cur->r) {
      cur->cnt = 0;
      return;
    }
    if (cur->lson->mxr == cur->rson->mxl - 1)
      cur->cnt = cur->lson->cnt + cur->rson->cnt - 1;
    else
      cur->cnt = cur->lson->cnt + cur->rson->cnt;
    if (!cur->lson->mxl) {
      if (cur->rson->mxl)
        cur->mxl = cur->rson->mxl;
      else
        cur->mxl = 0;
    } else
      cur->mxl = cur->lson->mxl;
    if (!cur->rson->mxr) {
      if (cur->lson->mxr)
        cur->mxr = cur->lson->mxr;
      else
        cur->mxr = 0;
    } else
      cur->mxr = cur->rson->mxr;
  }
}
void build(node *cur, int l, int r) {
  cur->l = l, cur->r = r;
  cur->cover = cur->cnt = 0;
  cur->mxl = cur->mxr = 0;
  if (l == r) return;
  int mid = l + r >> 1;
  build(Newnode(cur->lson), l, mid);
  build(Newnode(cur->rson), mid + 1, r);
}
void modify(node *cur, int l, int r, int v) {
  if (r < l) l = r;
  if (cur->l == l && cur->r == r) {
    cur->cover += v;
    update(cur);
    return;
  }
  int mid = cur->l + cur->r >> 1;
  if (r <= mid)
    modify(cur->lson, l, r, v);
  else if (l > mid)
    modify(cur->rson, l, r, v);
  else
    modify(cur->lson, l, mid, v), modify(cur->rson, mid + 1, r, v);
  update(cur);
}
void Discretization() {
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++)
    if (!mapp[v[i]]) cnt++, mapp[v[i]] = cnt;
}
int main() {
  read(T);
  while (T--) {
    read(n);
    for (int i = 1; i <= n; i++)
      read(s[i].l), read(s[i].r), v.push_back(s[i].l), v.push_back(s[i].r);
    tot = cnt = 0;
    Discretization();
    build(Newnode(root), 1, 2 * cnt);
    for (int i = 1; i <= n; i++)
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, 1);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, -1);
      ans = max(ans, root->cnt);
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, 1);
    }
    printf("%d\n", ans);
    v.clear(), mapp.clear();
  }
  return 0;
}

### 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 int maxn = 2e5 + 6;
int T;
int n, tot, x, cnt;
map<int, int> mapp;
vector<int> v;
template <class T>
void read(T &x, int f = 1) {
  char ch = getchar();
  x = 0;
  while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : 1), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
struct node {
  node *lson, *rson;
  int cnt, cover, l, r;
  int mxl, mxr;
} tr[maxn * 8], *root;
struct segment {
  int l, r;
} s[maxn];
inline node *Newnode(node *&root) { return root = &tr[tot++]; }
inline void update(node *cur) {
  if (cur->cover)
    cur->cnt = 1, cur->mxl = cur->l, cur->mxr = cur->r;
  else {
    cur->mxl = cur->mxr = 0;
    if (cur->l == cur->r) {
      cur->cnt = 0;
      return;
    }
    if (cur->lson->mxr == cur->rson->mxl - 1)
      cur->cnt = cur->lson->cnt + cur->rson->cnt - 1;
    else
      cur->cnt = cur->lson->cnt + cur->rson->cnt;
    if (!cur->lson->mxl) {
      if (cur->rson->mxl)
        cur->mxl = cur->rson->mxl;
      else
        cur->mxl = 0;
    } else
      cur->mxl = cur->lson->mxl;
    if (!cur->rson->mxr) {
      if (cur->lson->mxr)
        cur->mxr = cur->lson->mxr;
      else
        cur->mxr = 0;
    } else
      cur->mxr = cur->rson->mxr;
  }
}
void build(node *cur, int l, int r) {
  cur->l = l, cur->r = r;
  cur->cover = cur->cnt = 0;
  cur->mxl = cur->mxr = 0;
  if (l == r) return;
  int mid = l + r >> 1;
  build(Newnode(cur->lson), l, mid);
  build(Newnode(cur->rson), mid + 1, r);
}
void modify(node *cur, int l, int r, int v) {
  if (r < l) l = r;
  if (cur->l == l && cur->r == r) {
    cur->cover += v;
    update(cur);
    return;
  }
  int mid = cur->l + cur->r >> 1;
  if (r <= mid)
    modify(cur->lson, l, r, v);
  else if (l > mid)
    modify(cur->rson, l, r, v);
  else
    modify(cur->lson, l, mid, v), modify(cur->rson, mid + 1, r, v);
  update(cur);
}
void Discretization() {
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++)
    if (!mapp[v[i]]) cnt++, mapp[v[i]] = cnt;
}
int main() {
  read(T);
  while (T--) {
    read(n);
    for (int i = 1; i <= n; i++)
      read(s[i].l), read(s[i].r), v.push_back(s[i].l), v.push_back(s[i].r);
    tot = cnt = 0;
    Discretization();
    build(Newnode(root), 1, 2 * cnt);
    for (int i = 1; i <= n; i++)
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, 1);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, -1);
      ans = max(ans, root->cnt);
      modify(root, mapp[s[i].l] * 2, mapp[s[i].r] * 2 - 1, 1);
    }
    printf("%d\n", ans);
    v.clear(), mapp.clear();
  }
  return 0;
}

```