/**
 * Denis has N boxes (N even), and each box contains M marbles which are either 
 * black or white. He knows for each box how many black marbles it contains and 
 * how many white marbles it contains. He now wants to split the N boxes into 2 
 * groups, each containing N/2 boxes, such that the sum of the black marbles in 
 * the first group, and the white marbles in the second group is maximum.
 *
 * Input
 * The first line of input contains N and M, in this order.
 * The following N lines contain two positive integers each, denoting the number 
 * of white marbles and the number of black marbles that are in each box.
 *
 * Output
 * On a single line print two integers separated by space: the number of black 
 * marbles in the first group and the number of white marbles in the second group.
 *
 * Constraints
 * 1 ≤ N ≤ 105
 * 0 ≤ M ≤ 104
 * The solution is unique.
 *
 * Sample
 * Input    Output
 * 4 3       5 5
 * 0 3
 * 3 0
 * 1 2
 * 2 1
 *
 * TIME LIMIT = 3s
 */

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, m, i, j, temp, sum_white = 0, sum_black = 0;
    int **date;
   	scanf("%d %d", &n, &m);
    //read input
    date = (int **)malloc(n*sizeof(int*));
    for (i = 0; i<n; i++) {
        date[i] = (int *)malloc(2*sizeof(int));
        //read values
        scanf("%d %d",&date[i][0],&date[i][1]);
    }

    //need to sort date[][] by the first column
    for (i = 1; i < n; i++){
        for (j = 0; j < n-1; j++){
            if(date[j][0] < date[j+1][0]){
                temp = date[j][0];
                date[j][0] = date[j+1][0];
                date[j+1][0] = temp;
                temp = date[j][1];
                date[j][1] = date[j+1][1];
                date[j+1][1] = temp;
            }
        }
    }
    for (i=0;i<n;i++){
        if (i>(n/2-1)){
            sum_white += date[i][1];
        } else {
            sum_black += date[i][0];
        }
    }

    printf("%d %d\n",sum_black,sum_white);

    /*
    //free allocated memory
    for (i = 0; i<n; i++) {
        free(date[i]);
    }
    free(date);
    */
    return 0;
}