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
3
1 1
7 5
1 5
output
2
input
6
0 0
0 1
0 2
-1 1
0 1
1 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.
Source
Codeforces Round #345 (Div. 2) C. Watchmen
My Solution
build a
struct point{ int x, y; } pts[maxn];
to store the data, and sort with .x first and to calculate the same x that is on continuouslyand use C(2, t) = t*(t-1)/2 add to the sumx;
and sumy is the same way to do as the sumx did.
Then I use a map > cnt to count the same coordinates when input, and them use iterator to count the sumsame;
All of sumx, sumy, sumsame, t, num should be long long, because the t,num can be 2*1e5 so the C(2, t) and C(2, num) can be 400 1e8 large.
.
for Daniel. For pairs 
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