#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
int root[300001];
int find(int u) {
  if (root[u] != u) root[u] = find(root[u]);
  return root[u];
}
bool visited[300001];
int active[300001];
int par[300001];
vector<int> stk;
void dfs(int u, const vector<vector<int>>& ke) {
  visited[u] = true;
  root[u] = u;
  stk.push_back(u);
  for (int v : ke[u])
    if (visited[v]) {
      v = find(active[v]);
      while (stk.back() != v) {
        root[find(stk.back())] = v;
        stk.pop_back();
      }
    }
  for (int v : ke[u])
    if (!visited[v]) {
      active[u] = v;
      par[v] = u;
      dfs(v, ke);
    }
  if (stk.back() == u) stk.pop_back();
}
void chemthan() {
  int test;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    vector<pair<int, int>> vals(n);
    for (int i = (0); i < (n); ++i) cin >> vals[i].first >> vals[i].second;
    sort((vals).begin(), (vals).end());
    set<pair<int, int>> st;
    vector<vector<int>> adj(n);
    for (int i = (0); i < (int((vals).size())); ++i) {
      auto it = st.lower_bound(make_pair(vals[i].first, -INF));
      if (it != st.end()) {
        adj[i].push_back(it->second);
        adj[it->second].push_back(i);
      }
      it = st.end();
      if (it != st.begin()) {
        it--;
        if (vals[i].first <= it->first) {
          adj[i].push_back(it->second);
          adj[it->second].push_back(i);
        }
      }
      st.insert(make_pair(vals[i].second, i));
    }
    for (int u = (0); u < (n); ++u) {
      sort((adj[u]).begin(), (adj[u]).end()),
          (adj[u]).erase(unique((adj[u]).begin(), (adj[u]).end()),
                         (adj[u]).end());
    }
    int tot = 0;
    for (int u = (0); u < (n); ++u)
      root[u] = u, visited[u] = active[u] = 0, par[u] = -1;
    for (int u = (0); u < (n); ++u) {
      if (!visited[u]) {
        dfs(u, adj);
        tot++;
      }
    }
    int res = -1;
    for (int u = (0); u < (n); ++u) {
      map<int, int> hs;
      for (int v : adj[u]) {
        int t = find(v);
        if (par[t] != -1) {
          hs[par[t]] = 1;
        }
      }
      vector<int> vals;
      for (int v : adj[u])
        if (!hs[v]) {
          vals.push_back(find(v));
        }
      sort((vals).begin(), (vals).end()),
          (vals).erase(unique((vals).begin(), (vals).end()), (vals).end());
      chkmax(res, int((vals).size()) - 1);
    }
    cout << tot + res << "\n";
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
int root[300001];
int find(int u) {
  if (root[u] != u) root[u] = find(root[u]);
  return root[u];
}
bool visited[300001];
int active[300001];
int par[300001];
vector<int> stk;
void dfs(int u, const vector<vector<int>>& ke) {
  visited[u] = true;
  root[u] = u;
  stk.push_back(u);
  for (int v : ke[u])
    if (visited[v]) {
      v = find(active[v]);
      while (stk.back() != v) {
        root[find(stk.back())] = v;
        stk.pop_back();
      }
    }
  for (int v : ke[u])
    if (!visited[v]) {
      active[u] = v;
      par[v] = u;
      dfs(v, ke);
    }
  if (stk.back() == u) stk.pop_back();
}
void chemthan() {
  int test;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    vector<pair<int, int>> vals(n);
    for (int i = (0); i < (n); ++i) cin >> vals[i].first >> vals[i].second;
    sort((vals).begin(), (vals).end());
    set<pair<int, int>> st;
    vector<vector<int>> adj(n);
    for (int i = (0); i < (int((vals).size())); ++i) {
      auto it = st.lower_bound(make_pair(vals[i].first, -INF));
      if (it != st.end()) {
        adj[i].push_back(it->second);
        adj[it->second].push_back(i);
      }
      it = st.end();
      if (it != st.begin()) {
        it--;
        if (vals[i].first <= it->first) {
          adj[i].push_back(it->second);
          adj[it->second].push_back(i);
        }
      }
      st.insert(make_pair(vals[i].second, i));
    }
    for (int u = (0); u < (n); ++u) {
      sort((adj[u]).begin(), (adj[u]).end()),
          (adj[u]).erase(unique((adj[u]).begin(), (adj[u]).end()),
                         (adj[u]).end());
    }
    int tot = 0;
    for (int u = (0); u < (n); ++u)
      root[u] = u, visited[u] = active[u] = 0, par[u] = -1;
    for (int u = (0); u < (n); ++u) {
      if (!visited[u]) {
        dfs(u, adj);
        tot++;
      }
    }
    int res = -1;
    for (int u = (0); u < (n); ++u) {
      map<int, int> hs;
      for (int v : adj[u]) {
        int t = find(v);
        if (par[t] != -1) {
          hs[par[t]] = 1;
        }
      }
      vector<int> vals;
      for (int v : adj[u])
        if (!hs[v]) {
          vals.push_back(find(v));
        }
      sort((vals).begin(), (vals).end()),
          (vals).erase(unique((vals).begin(), (vals).end()), (vals).end());
      chkmax(res, int((vals).size()) - 1);
    }
    cout << tot + res << "\n";
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```