codeforces 651C Watchmen

C. Watchmen

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi - xj| + |yi - yj|. Daniel, as an ordinary person, calculates the distance using the formula .

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples

input

31 17 51 5

output

2

input

60 00 10 2-1 10 11 1

output

11

Note

In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and  for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

用三个map，mx记录一种x的数量，my记录一种y的数量，mp用来记录一种点对的数量

对于相同x的即mx[x] == c1，则它在x相同的的这些点对里的组合情况是(c1 - 1) + (c1 - 2) + (c1 - 3) + ... + 1 = c1 * (c1 - 1) / 2种

对于相同y的即my[y] == c2，则它在y相同的这些点对里的组合情况是(c2 - 1) + (c2 - 2) + (c2 - 3) + ... + 1 = c2 * (c2 - 1) / 2种

但是这么算可能会有重复的即对于例如(a，b），(a，b)，(a，b）这三个点，横纵坐标都一样，也就是说对于相同的x它们内部会算一遍，对于相同的y它们内部之间也会算一遍，这时候就需要mp[pair(a,b)] == c3了，用于这三个(a, b)内部去重，要减去(c3 - 1) + (c3 - 2) + ... + 1 = c3 * (c3 - 1)

#include <cstdio>#include <cstring>#include <iostream>#include <map>using namespace std;typedef long long ll;typedef pair<int, int> p;map<int, int> mx, my;map<p, int> mp;int main(){int n, x, y;scanf("%d", &n);mx.clear();my.clear();mp.clear();for (int k = 1; k <= n; k++) {scanf("%d%d", &x, &y);mx[x]++; my[y]++;mp[p(x, y)]++;}ll ans = 0;map<int, int>::iterator i;int tem;for (i = mx.begin(); i != mx.end(); i++) {tem = i->second;ans += (ll)tem * (tem - 1) / 2;}for (i = my.begin(); i != my.end(); i++) {tem = i->second;ans += (ll)tem * (tem - 1) / 2;}for (map<p, int>::iterator j = mp.begin(); j != mp.end(); j++) {tem = j->second;ans -= (ll)tem * (tem - 1) / 2;}printf("%I64d\n", ans);return 0;}