The Morning Star晨星

time limit per test

2 seconds

memory limit per test

256 megabytes

A compass points directly toward the morning star. It can only point in one of eight directions: the four cardinal directions (N, S, E, W) or some combination (NW, NE, SW, SE). Otherwise, it will break.

The directions the compass can point.

There are nn distinct points with integer coordinates on a plane. How many ways can you put a compass at one point and the morning star at another so that the compass does not break?

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104). The description of the test cases follows.

The first line of each test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) --- the number of points.

Then nn lines follow, each line containing two integers xixi, yiyi (−109≤xi,yi≤109−109≤xi,yi≤109) --- the coordinates of each point, all points have distinct coordinates.

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

For each test case, output a single integer --- the number of pairs of points that don't break the compass.

Example

Input

Copy

复制代码

5

3

0 0

-1 -1

1 1

4

4 5

5 7

6 9

10 13

3

-1000000000 1000000000

0 0

1000000000 -1000000000

5

0 0

2 2

-1 5

-1 10

2 11

3

0 0

-1 2

1 -2

Output

Copy

复制代码
6
2
6
8
0

Note

In the first test case, any pair of points won't break the compass:

  • The compass is at (0,0)(0,0), the morning star is at (−1,−1)(−1,−1): the compass will point SWSW.
  • The compass is at (0,0)(0,0), the morning star is at (1,1)(1,1): the compass will point NENE.
  • The compass is at (−1,−1)(−1,−1), the morning star is at (0,0)(0,0): the compass will point NENE.
  • The compass is at (−1,−1)(−1,−1), the morning star is at (1,1)(1,1): the compass will point NENE.
  • The compass is at (1,1)(1,1), the morning star is at (0,0)(0,0): the compass will point SWSW.
  • The compass is at (1,1)(1,1), the morning star is at (−1,−1)(−1,−1): the compass will point SWSW.

In the second test case, only two pairs of points won't break the compass:

  • The compass is at (6,9)(6,9), the morning star is at (10,13)(10,13): the compass will point NENE.
  • The compass is at (10,13)(10,13), the morning star is at (6,9)(6,9): the compass will point SWSW.

每个测试的时间限制 2 秒

每个测试的内存限制 256 兆字节

指南针直接指向晨星。它只能指向八个方向之一:四个基本方向(N、S、E、W)或某种组合(NW、NE、SW、SE)。否则,它会损坏。

指南针可以指向的方向。

平面上有 n

个具有整数坐标的不同点。有多少种方法可以将指南针放在一个点,将晨星放在另一个点,这样指南针就不会损坏?

输入

每个测试包含多个测试用例。第一行包含测试用例数 t

(1≤t≤104

)。测试用例的描述如下。

每个测试用例的第一行包含一个整数 n

(2≤n≤2⋅105

)--- 点数。

接下来是 n

行,每行包含两个整数 xi

, yi

(−109≤xi,yi≤109

) --- 每个点的坐标,所有点都有不同的坐标。

保证所有测试用例的 n

之和不超过 2⋅105

输出

对于每个测试用例,输出一个整数 --- 不偏离指南针的点对的数量。

示例

输入副本

5

3

0 0

-1 -1

1 1

4

4 5

5 7

6 9

10 13

3

-1000000000 1000000000

0 0

1000000000 -1000000000

5

0 0

2 2

-1 5

-1 10

2 11

3

0 0

-1 2

1 -2

输出副本

6

2

6

8

0

注意

在第一个测试用例中,任何一对点都不会破坏指南针:

指南针位于 (0,0)

,晨星位于 (-1,-1)

:指南针将指向 SW

指南针位于 (0,0)

,晨星位于 (1,1)

:指南针将指向 NE

罗盘位于 (−1,−1)

,晨星位于 (0,0)

:罗盘将指向东北

罗盘位于 (−1,−1)

,晨星位于 (1,1)

:罗盘将指向东北

罗盘位于 (1,1)

,晨星位于 (0,0)

:罗盘将指向西南

罗盘位于 (1,1)

,晨星位于 (−1,−1)

:罗盘将指向西南

在第二个测试案例中,只有两对点不会破坏罗盘:

罗盘位于 (6,9)

,晨星位于 (10,13)

:罗盘将指向东北

罗盘位于 (10,13)

,晨星位于 (6,9)

:罗盘将指向西南

代码:

cpp 复制代码
#include <iostream>
#include <vector>
#include <map>
using namespace std;

#define ll long long

ll A2(ll m) {
    return m * (m - 1);
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<pair<ll, ll>> points(n);
        map<ll, ll> x_count, y_count, x_add_y_count, x_sub_y_count;

        for (int i = 0; i < n; i++) {
            cin >> points[i].first >> points[i].second;
            x_count[points[i].first]++;
            y_count[points[i].second]++;
            x_add_y_count[points[i].first + points[i].second]++;
            x_sub_y_count[points[i].first - points[i].second]++;
        }

        ll ans = 0;

        for (auto& kv : x_count) {
            ans += A2(kv.second);
        }
        for (auto& kv : y_count) {
            ans += A2(kv.second);
        }
        for (auto& kv : x_add_y_count) {
            ans += A2(kv.second);
        }
        for (auto& kv : x_sub_y_count) {
            ans += A2(kv.second);
        }

        cout << ans << endl;
    }
    return 0;
}
相关推荐
vibecoding日记2 小时前
双非如何快速入职字节等大厂大模型?真实案例分析:推理优化和投机解码
算法·求职·大模型工程师
yszaygr21384 小时前
Verilog参数化游程编码RLE模块
算法
望易4 小时前
刚设计的大模型架构-双域耦合认知框架
算法·架构
复杂网络8 小时前
多个 Claude Code 与多个 Codex 协同工作:设计与实现方案
算法
apocelipes1 天前
常用编程语言和库的正则表达式性能对比
c语言·c++·python·性能优化·golang·开发工具和环境
HjhIron1 天前
面试常客:字符串算法从入门到进阶
算法·面试
吴佳浩1 天前
DeepSeek DSpark:Confidence-Scheduled Speculative Decoding 技术解析
人工智能·算法·deepseek
触底反弹1 天前
🧠 搞懂 Token,才算真正入门大模型——从分词原理到 Embedding 语义实战
javascript·人工智能·算法
vivo互联网技术1 天前
ICLR 2026 | 基于后验采样的图像恢复方法LearnIR:人脸去阴影、去雾
人工智能·算法·aigc
浮生望1 天前
JS字符串与回文算法:从包装类到双指针的面试进阶之路
javascript·算法