#include <bits/stdc++.h>
using namespace std;
const long long BINF = 9e18, LINF = 2e9, P = 179, Q = 1791791791;
const long long MAXN = 2e5;
map<long long, long long> compress(vector<long long> a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  map<long long, long long> make_pair;
  for (long long i = 0; i < a.size(); ++i) make_pair[a[i]] = i;
  return make_pair;
}
long long l[MAXN], r[MAXN];
pair<long long, long long> t1[MAXN * 8], t2[MAXN * 8];
long long psh1[MAXN * 8], psh2[MAXN * 8];
void build(pair<long long, long long>* t, long long* psh, long long v,
           long long tl, long long tr) {
  if (tl == tr)
    t[v] = {0ll, 1ll}, psh[v] = 0;
  else {
    long long tm = (tl + tr) / 2;
    build(t, psh, v * 2, tl, tm);
    build(t, psh, v * 2 + 1, tm + 1, tr);
    t[v] = {0ll, tr - tl + 1};
    psh[v] = 0;
  }
}
void push(long long* psh, pair<long long, long long>* t, long long v,
          long long tl, long long tr) {
  if (tl != tr) {
    psh[v * 2] += psh[v];
    psh[v * 2 + 1] += psh[v];
  }
  t[v].first += psh[v];
  psh[v] = 0;
}
void recalc(pair<long long, long long>* t, long long v) {
  if (t[v * 2].first == t[2 * v + 1].first)
    t[v] = {t[v * 2].first, t[v * 2].second + t[v * 2 + 1].second};
  else
    t[v] = min(t[v * 2], t[v * 2 + 1]);
}
void add(pair<long long, long long>* t, long long* psh, long long v,
         long long tl, long long tr, long long l, long long r, long long d) {
  push(psh, t, v, tl, tr);
  if (r < tl || tr < l) return;
  if (l <= tl && tr <= r) {
    psh[v] += d;
    push(psh, t, v, tl, tr);
    return;
  }
  long long tm = (tl + tr) / 2;
  add(t, psh, v * 2, tl, tm, l, r, d);
  add(t, psh, v * 2 + 1, tm + 1, tr, l, r, d);
  recalc(t, v);
}
long long C;
long long f() { return (C + 1 - t1[1].second) - (C + 1 - t2[1].second); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> v;
  for (long long i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
    v.push_back(l[i]);
    v.push_back(r[i]);
  }
  map<long long, long long> crd = compress(v);
  for (long long i = 0; i < n; ++i) l[i] = crd[l[i]], r[i] = crd[r[i]];
  C = crd.size();
  build(t1, psh1, 1, 0, C);
  build(t2, psh2, 1, 0, C);
  for (long long i = 0; i < n; ++i) {
    add(t1, psh1, 1, 0, C, l[i], r[i], 1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, 1);
  }
  long long ans = -1;
  for (long long i = 0; i < n; ++i) {
    add(t1, psh1, 1, 0, C, l[i], r[i], -1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, -1);
    ans = max(ans, f());
    add(t1, psh1, 1, 0, C, l[i], r[i], 1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, 1);
  }
  cout << ans << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout.precision(40);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; ++i) solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long BINF = 9e18, LINF = 2e9, P = 179, Q = 1791791791;
const long long MAXN = 2e5;
map<long long, long long> compress(vector<long long> a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  map<long long, long long> make_pair;
  for (long long i = 0; i < a.size(); ++i) make_pair[a[i]] = i;
  return make_pair;
}
long long l[MAXN], r[MAXN];
pair<long long, long long> t1[MAXN * 8], t2[MAXN * 8];
long long psh1[MAXN * 8], psh2[MAXN * 8];
void build(pair<long long, long long>* t, long long* psh, long long v,
           long long tl, long long tr) {
  if (tl == tr)
    t[v] = {0ll, 1ll}, psh[v] = 0;
  else {
    long long tm = (tl + tr) / 2;
    build(t, psh, v * 2, tl, tm);
    build(t, psh, v * 2 + 1, tm + 1, tr);
    t[v] = {0ll, tr - tl + 1};
    psh[v] = 0;
  }
}
void push(long long* psh, pair<long long, long long>* t, long long v,
          long long tl, long long tr) {
  if (tl != tr) {
    psh[v * 2] += psh[v];
    psh[v * 2 + 1] += psh[v];
  }
  t[v].first += psh[v];
  psh[v] = 0;
}
void recalc(pair<long long, long long>* t, long long v) {
  if (t[v * 2].first == t[2 * v + 1].first)
    t[v] = {t[v * 2].first, t[v * 2].second + t[v * 2 + 1].second};
  else
    t[v] = min(t[v * 2], t[v * 2 + 1]);
}
void add(pair<long long, long long>* t, long long* psh, long long v,
         long long tl, long long tr, long long l, long long r, long long d) {
  push(psh, t, v, tl, tr);
  if (r < tl || tr < l) return;
  if (l <= tl && tr <= r) {
    psh[v] += d;
    push(psh, t, v, tl, tr);
    return;
  }
  long long tm = (tl + tr) / 2;
  add(t, psh, v * 2, tl, tm, l, r, d);
  add(t, psh, v * 2 + 1, tm + 1, tr, l, r, d);
  recalc(t, v);
}
long long C;
long long f() { return (C + 1 - t1[1].second) - (C + 1 - t2[1].second); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> v;
  for (long long i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
    v.push_back(l[i]);
    v.push_back(r[i]);
  }
  map<long long, long long> crd = compress(v);
  for (long long i = 0; i < n; ++i) l[i] = crd[l[i]], r[i] = crd[r[i]];
  C = crd.size();
  build(t1, psh1, 1, 0, C);
  build(t2, psh2, 1, 0, C);
  for (long long i = 0; i < n; ++i) {
    add(t1, psh1, 1, 0, C, l[i], r[i], 1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, 1);
  }
  long long ans = -1;
  for (long long i = 0; i < n; ++i) {
    add(t1, psh1, 1, 0, C, l[i], r[i], -1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, -1);
    ans = max(ans, f());
    add(t1, psh1, 1, 0, C, l[i], r[i], 1);
    if (l[i] != r[i]) add(t2, psh2, 1, 0, C, l[i], r[i] - 1, 1);
  }
  cout << ans << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout.precision(40);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; ++i) solve();
  return 0;
}

```