#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 201000;
int l[MAX_N], r[MAX_N];
int sub_a[2 * MAX_N];
int b[4 * MAX_N], sum[4 * MAX_N];
struct node {
  int l, r, val;
  int vl, vr;
} a[MAX_N * 16];
void update(int k) {
  a[k].vl = a[k << 1].vl;
  a[k].vr = a[k << 1 | 1].vr;
  if (a[k << 1].vr == 1 && a[k << 1 | 1].vl == 1)
    a[k].val = a[k << 1].val + a[k << 1 | 1].val - 1;
  else
    a[k].val = a[k << 1].val + a[k << 1 | 1].val;
}
void build(int k, int l, int r) {
  a[k].l = l;
  a[k].r = r;
  a[k].val = 0;
  if (l == r) {
    a[k].vl = sum[l];
    a[k].vr = sum[r];
    if (sum[l] == 1) a[k].val = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  update(k);
}
int query(int k, int l, int r) {
  if (a[k].l >= l && a[k].r <= r) return a[k].val;
  int mid = (a[k].l + a[k].r) >> 1;
  int x = 0;
  if (r > mid) x += query(k << 1 | 1, l, r);
  if (l <= mid) x += query(k << 1, l, r);
  if (r > mid && l <= mid && a[k << 1].vr == 1 && a[k << 1 | 1].vl == 1) x--;
  return x;
}
int main(void) {
  int T, n, i;
  cin >> T;
  while (T--) {
    scanf("%d", &n);
    int tot = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d%d", &l[i], &r[i]);
      sub_a[tot++] = l[i];
      sub_a[tot++] = r[i];
    }
    sort(sub_a, sub_a + tot);
    int nn = unique(sub_a, sub_a + tot) - sub_a;
    for (i = 1; i <= n; i++) {
      l[i] = lower_bound(sub_a, sub_a + nn, l[i]) - sub_a + 1;
      r[i] = lower_bound(sub_a, sub_a + nn, r[i]) - sub_a + 1;
    }
    int k = 4 * n + 10;
    for (i = 1; i <= k; i++) {
      b[i] = sum[i] = 0;
    }
    for (i = 1; i <= n; i++) {
      b[2 * l[i]]++;
      b[2 * r[i] + 1]--;
    }
    for (i = 1; i <= k; i++) {
      sum[i] = sum[i - 1] + b[i];
    }
    for (i = 1; i <= k; i++) {
      if (sum[i] >= 2) sum[i] = 2;
    }
    int ans = 1;
    int flag = 1;
    int t = 0;
    for (i = 1; i <= k; i++) {
      if (sum[i] != 0 && flag) {
        t++;
        flag = 0;
      } else if (sum[i] == 0)
        flag = 1;
    }
    build(1, 1, k);
    for (i = 1; i <= n; i++) {
      int res = query(1, 2 * l[i], 2 * r[i]);
      if (sum[2 * l[i]] == 1 && sum[2 * l[i] - 1] == 0) res--;
      if (sum[2 * r[i]] == 1 && sum[2 * r[i] + 1] == 0) res--;
      res += t;
      ans = max(ans, res);
    }
    printf("%d\n", ans);
  }
  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 MAX_N = 201000;
int l[MAX_N], r[MAX_N];
int sub_a[2 * MAX_N];
int b[4 * MAX_N], sum[4 * MAX_N];
struct node {
  int l, r, val;
  int vl, vr;
} a[MAX_N * 16];
void update(int k) {
  a[k].vl = a[k << 1].vl;
  a[k].vr = a[k << 1 | 1].vr;
  if (a[k << 1].vr == 1 && a[k << 1 | 1].vl == 1)
    a[k].val = a[k << 1].val + a[k << 1 | 1].val - 1;
  else
    a[k].val = a[k << 1].val + a[k << 1 | 1].val;
}
void build(int k, int l, int r) {
  a[k].l = l;
  a[k].r = r;
  a[k].val = 0;
  if (l == r) {
    a[k].vl = sum[l];
    a[k].vr = sum[r];
    if (sum[l] == 1) a[k].val = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  update(k);
}
int query(int k, int l, int r) {
  if (a[k].l >= l && a[k].r <= r) return a[k].val;
  int mid = (a[k].l + a[k].r) >> 1;
  int x = 0;
  if (r > mid) x += query(k << 1 | 1, l, r);
  if (l <= mid) x += query(k << 1, l, r);
  if (r > mid && l <= mid && a[k << 1].vr == 1 && a[k << 1 | 1].vl == 1) x--;
  return x;
}
int main(void) {
  int T, n, i;
  cin >> T;
  while (T--) {
    scanf("%d", &n);
    int tot = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d%d", &l[i], &r[i]);
      sub_a[tot++] = l[i];
      sub_a[tot++] = r[i];
    }
    sort(sub_a, sub_a + tot);
    int nn = unique(sub_a, sub_a + tot) - sub_a;
    for (i = 1; i <= n; i++) {
      l[i] = lower_bound(sub_a, sub_a + nn, l[i]) - sub_a + 1;
      r[i] = lower_bound(sub_a, sub_a + nn, r[i]) - sub_a + 1;
    }
    int k = 4 * n + 10;
    for (i = 1; i <= k; i++) {
      b[i] = sum[i] = 0;
    }
    for (i = 1; i <= n; i++) {
      b[2 * l[i]]++;
      b[2 * r[i] + 1]--;
    }
    for (i = 1; i <= k; i++) {
      sum[i] = sum[i - 1] + b[i];
    }
    for (i = 1; i <= k; i++) {
      if (sum[i] >= 2) sum[i] = 2;
    }
    int ans = 1;
    int flag = 1;
    int t = 0;
    for (i = 1; i <= k; i++) {
      if (sum[i] != 0 && flag) {
        t++;
        flag = 0;
      } else if (sum[i] == 0)
        flag = 1;
    }
    build(1, 1, k);
    for (i = 1; i <= n; i++) {
      int res = query(1, 2 * l[i], 2 * r[i]);
      if (sum[2 * l[i]] == 1 && sum[2 * l[i] - 1] == 0) res--;
      if (sum[2 * r[i]] == 1 && sum[2 * r[i] + 1] == 0) res--;
      res += t;
      ans = max(ans, res);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```