#include <bits/stdc++.h>
using namespace std;
const int __ = 2e5 + 5;
struct node {
  int l, r;
} a[__];
double b[__ << 2];
struct SegmentTree {
  int n, ql, qr, val;
  struct node {
    int v, c;
    bool lv, rv;
    void putadd(int val) { c += val; }
  } t[__ << 5];
  void pushup(int x) {
    if (t[x].c) {
      t[x] = {1, t[x].c, true, true};
      return;
    }
    t[x].v = t[x << 1].v + t[x << 1 | 1].v;
    if (t[x << 1].rv && t[x << 1 | 1].lv) --t[x].v;
    t[x].lv = t[x << 1].lv;
    t[x].rv = t[x << 1 | 1].rv;
  }
  void build(int _n) {
    n = _n;
    build(1, 1, n);
  }
  void build(int x, int tl, int tr) {
    t[x] = {0, 0, false, false};
    t[x << 1] = {0, 0, false, false};
    t[x << 1 | 1] = {0, 0, false, false};
    if (tl == tr) return;
    int tm = (tl + tr) >> 1;
    build(x << 1, tl, tm);
    build(x << 1 | 1, tm + 1, tr);
    pushup(x);
  }
  void add(int _ql, int _qr, int v) {
    ql = _ql, qr = _qr, val = v;
    _add(1, 1, n);
  }
  void _add(int x, int tl, int tr) {
    if (ql > tr || qr < tl) return;
    if (ql <= tl && tr <= qr) {
      t[x].putadd(val);
      pushup(x);
      return;
    }
    int tm = (tl + tr) >> 1;
    _add(x << 1, tl, tm);
    _add(x << 1 | 1, tm + 1, tr);
    pushup(x);
  }
  int get() { return t[1].v; }
} T;
int main() {
  int _;
  for (scanf("%d", &_); _; --_) {
    int n, m = 0;
    scanf("%d", &n);
    for (int i = (1); i <= (n); ++i) {
      scanf("%d%d", &a[i].l, &a[i].r);
      b[++m] = a[i].l - 0.3;
      b[++m] = a[i].l;
      b[++m] = a[i].r;
      b[++m] = a[i].r + 0.3;
    }
    sort(b + 1, b + 1 + m);
    m = unique(b + 1, b + 1 + m) - b - 1;
    T.build(m);
    for (int i = 1; i <= n; ++i) {
      a[i].l = lower_bound(b + 1, b + 1 + m, a[i].l) - b;
      a[i].r = lower_bound(b + 1, b + 1 + m, a[i].r) - b;
      T.add(a[i].l, a[i].r, 1);
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      T.add(a[i].l, a[i].r, -1);
      ans = max(ans, T.get());
      T.add(a[i].l, a[i].r, 1);
    }
    printf("%d\n", ans);
  }
  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 int __ = 2e5 + 5;
struct node {
  int l, r;
} a[__];
double b[__ << 2];
struct SegmentTree {
  int n, ql, qr, val;
  struct node {
    int v, c;
    bool lv, rv;
    void putadd(int val) { c += val; }
  } t[__ << 5];
  void pushup(int x) {
    if (t[x].c) {
      t[x] = {1, t[x].c, true, true};
      return;
    }
    t[x].v = t[x << 1].v + t[x << 1 | 1].v;
    if (t[x << 1].rv && t[x << 1 | 1].lv) --t[x].v;
    t[x].lv = t[x << 1].lv;
    t[x].rv = t[x << 1 | 1].rv;
  }
  void build(int _n) {
    n = _n;
    build(1, 1, n);
  }
  void build(int x, int tl, int tr) {
    t[x] = {0, 0, false, false};
    t[x << 1] = {0, 0, false, false};
    t[x << 1 | 1] = {0, 0, false, false};
    if (tl == tr) return;
    int tm = (tl + tr) >> 1;
    build(x << 1, tl, tm);
    build(x << 1 | 1, tm + 1, tr);
    pushup(x);
  }
  void add(int _ql, int _qr, int v) {
    ql = _ql, qr = _qr, val = v;
    _add(1, 1, n);
  }
  void _add(int x, int tl, int tr) {
    if (ql > tr || qr < tl) return;
    if (ql <= tl && tr <= qr) {
      t[x].putadd(val);
      pushup(x);
      return;
    }
    int tm = (tl + tr) >> 1;
    _add(x << 1, tl, tm);
    _add(x << 1 | 1, tm + 1, tr);
    pushup(x);
  }
  int get() { return t[1].v; }
} T;
int main() {
  int _;
  for (scanf("%d", &_); _; --_) {
    int n, m = 0;
    scanf("%d", &n);
    for (int i = (1); i <= (n); ++i) {
      scanf("%d%d", &a[i].l, &a[i].r);
      b[++m] = a[i].l - 0.3;
      b[++m] = a[i].l;
      b[++m] = a[i].r;
      b[++m] = a[i].r + 0.3;
    }
    sort(b + 1, b + 1 + m);
    m = unique(b + 1, b + 1 + m) - b - 1;
    T.build(m);
    for (int i = 1; i <= n; ++i) {
      a[i].l = lower_bound(b + 1, b + 1 + m, a[i].l) - b;
      a[i].r = lower_bound(b + 1, b + 1 + m, a[i].r) - b;
      T.add(a[i].l, a[i].r, 1);
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      T.add(a[i].l, a[i].r, -1);
      ans = max(ans, T.get());
      T.add(a[i].l, a[i].r, 1);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```