B. Matchmaker

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus has n markers and m marker caps. Each marker is described by two numbers: x_i is the color and y_i is the diameter. Correspondingly, each cap is described by two numbers: a_j is the color and b_j is the diameter. Cap (a_j, b_j) can close marker (x_i, y_i) only if their diameters match, that is, b_j = y_i. Besides, a marker is considered to be beautifully closed, if the cap color and the marker color match, that is, a_j = x_i.

Find the way to close the maximum number of markers. If there are several such ways, then choose the one that has the maximum number of beautifully closed markers.

Input

The first input line contains two space-separated integers n and m (1 ≤ n, m ≤ 10⁵) — the number of markers and the number of caps, correspondingly.

Next n lines describe the markers. The i-th line contains two space-separated integers x_i, y_i (1 ≤ x_i, y_i ≤ 1000) — the i-th marker's color and diameter, correspondingly.

Next m lines describe the caps. The j-th line contains two space-separated integers a_j, b_j (1 ≤ a_j, b_j ≤ 1000) — the color and diameter of the j-th cap, correspondingly.

Output

Print two space-separated integers u, v, where u is the number of closed markers and v is the number of beautifully closedmarkers in the sought optimal way. Remember that you have to find the way to close the maximum number of markers, and if there are several such ways, you should choose the one where the number of beautifully closed markers is maximum.

Sample test(s)

input

3 41 23 42 45 42 41 11 2

output

3 2

input

2 21 22 13 45 1

output

1 0

Note

In the first test sample the first marker should be closed by the fourth cap, the second marker should be closed by the first cap and the third marker should be closed by the second cap. Thus, three markers will be closed, and two of them will bebeautifully closed — the first and the third markers.

解题说明：题目中提到了maker和maker cap两种东西，应该是用来匹配用的，此题要求的是closed makers和beautifully closed makers，其中beautiful的要求是除了半径要符合要求外，颜色也要符合要求。由于范围不是很大，可以用数组来进行标记，先读取maker信息在数组对应位置进行标记，再读入caps信息的时查看当前位置是否有maker已经标记。先判断半径，再判断颜色。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int main(){int v[1020][1020];int c[1020],n,m,x,y,a,b,i;    scanf("%d %d",&n,&m);    for(i=0;i<n;i++)    { scanf("%d %d",&x,&y);c[y]++;v[x][y]++;}    for(i=0;i<m;i++)    {        scanf("%d %d",&x,&y);        if(v[x][y])        {b++;v[x][y]--;}        if(c[y])         { a++; c[y]--;}    }    printf("%d %d\n",a,b);    return 0;}