#include <bits/stdc++.h>
const int MaxN = 8e5 + 50;
template <class t>
inline void read(t &s) {
  s = 0;
  register int f = 1;
  register char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  s *= f;
  return;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0) putchar('-'), x = -x;
  int buf[21], top = 0;
  while (x) buf[++top] = x % 10, x /= 10;
  if (!top) buf[++top] = 0;
  while (top) putchar(buf[top--] ^ '0');
  return;
}
inline void setIn(std::string s) {
  freopen(s.c_str(), "r", stdin);
  return;
}
inline void setOut(std::string s) {
  freopen(s.c_str(), "w", stdout);
  return;
}
inline void setIO(std::string s = "") {
  setIn(s + ".in");
  setOut(s + ".out");
  return;
}
template <class t>
inline bool ckmin(t &x, t y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
inline bool ckmax(t &x, t y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
inline int lowbit(int x) { return x & (-x); }
int l[MaxN], r[MaxN], n;
int f[MaxN], g[MaxN];
inline void work() {
  read(n);
  std::vector<int> dr;
  for (int i = 1; i <= n; ++i)
    read(l[i], r[i]), dr.push_back(l[i]), dr.push_back(r[i]);
  std::sort((dr).begin(), (dr).end()),
      dr.erase(std::unique((dr).begin(), (dr).end()), dr.end());
  for (int i = 1; i <= n; ++i)
    l[i] = std::lower_bound((dr).begin(), (dr).end(), l[i]) - dr.begin() + 1,
    r[i] = std::lower_bound((dr).begin(), (dr).end(), r[i]) - dr.begin() + 1;
  for (int i = 1; i <= n; ++i) l[i] = l[i] * 2 - 1, r[i] = r[i] * 2 - 1;
  register int m = *std::max_element(r + 1, r + n + 1);
  std::memset(f, 0, sizeof(int) * (m + 5)),
      std::memset(g, 0, sizeof(int) * (m + 5));
  for (int i = 1; i <= n; ++i) ++f[l[i]], --f[r[i] + 1];
  for (int i = 1; i <= m + 1; ++i) f[i] += f[i - 1];
  register int all = 0;
  {
    register int flg = 0;
    for (int i = 1; i <= m; ++i)
      if (f[i])
        flg = 1;
      else
        all += flg, flg = 0;
    all += flg;
  }
  {
    register int flg = 0;
    for (int i = 1; i <= m; ++i)
      if (f[i] == 1)
        flg = 1;
      else
        g[i - 1] = flg, flg = 0;
    g[m] = flg;
  }
  for (int i = 1; i <= m + 1; ++i) g[i] += g[i - 1];
  register int ans = -514114;
  for (int i = 1; i <= n; ++i)
    ckmax(ans, g[r[i] - 1] - g[l[i] - 1] + (f[r[i]] == 1 && f[r[i] + 1] >= 1) -
                   (!f[l[i] - 1] && f[l[i]] == 1));
  write(ans + all), std::puts("");
}
signed main(void) {
  int t;
  read(t);
  while (t--) work();
  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>
const int MaxN = 8e5 + 50;
template <class t>
inline void read(t &s) {
  s = 0;
  register int f = 1;
  register char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  s *= f;
  return;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0) putchar('-'), x = -x;
  int buf[21], top = 0;
  while (x) buf[++top] = x % 10, x /= 10;
  if (!top) buf[++top] = 0;
  while (top) putchar(buf[top--] ^ '0');
  return;
}
inline void setIn(std::string s) {
  freopen(s.c_str(), "r", stdin);
  return;
}
inline void setOut(std::string s) {
  freopen(s.c_str(), "w", stdout);
  return;
}
inline void setIO(std::string s = "") {
  setIn(s + ".in");
  setOut(s + ".out");
  return;
}
template <class t>
inline bool ckmin(t &x, t y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
inline bool ckmax(t &x, t y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
inline int lowbit(int x) { return x & (-x); }
int l[MaxN], r[MaxN], n;
int f[MaxN], g[MaxN];
inline void work() {
  read(n);
  std::vector<int> dr;
  for (int i = 1; i <= n; ++i)
    read(l[i], r[i]), dr.push_back(l[i]), dr.push_back(r[i]);
  std::sort((dr).begin(), (dr).end()),
      dr.erase(std::unique((dr).begin(), (dr).end()), dr.end());
  for (int i = 1; i <= n; ++i)
    l[i] = std::lower_bound((dr).begin(), (dr).end(), l[i]) - dr.begin() + 1,
    r[i] = std::lower_bound((dr).begin(), (dr).end(), r[i]) - dr.begin() + 1;
  for (int i = 1; i <= n; ++i) l[i] = l[i] * 2 - 1, r[i] = r[i] * 2 - 1;
  register int m = *std::max_element(r + 1, r + n + 1);
  std::memset(f, 0, sizeof(int) * (m + 5)),
      std::memset(g, 0, sizeof(int) * (m + 5));
  for (int i = 1; i <= n; ++i) ++f[l[i]], --f[r[i] + 1];
  for (int i = 1; i <= m + 1; ++i) f[i] += f[i - 1];
  register int all = 0;
  {
    register int flg = 0;
    for (int i = 1; i <= m; ++i)
      if (f[i])
        flg = 1;
      else
        all += flg, flg = 0;
    all += flg;
  }
  {
    register int flg = 0;
    for (int i = 1; i <= m; ++i)
      if (f[i] == 1)
        flg = 1;
      else
        g[i - 1] = flg, flg = 0;
    g[m] = flg;
  }
  for (int i = 1; i <= m + 1; ++i) g[i] += g[i - 1];
  register int ans = -514114;
  for (int i = 1; i <= n; ++i)
    ckmax(ans, g[r[i] - 1] - g[l[i] - 1] + (f[r[i]] == 1 && f[r[i] + 1] >= 1) -
                   (!f[l[i] - 1] && f[l[i]] == 1));
  write(ans + all), std::puts("");
}
signed main(void) {
  int t;
  read(t);
  while (t--) work();
  return 0;
}

```