#include<stdio.h>
#include<math.h>
double a[1000][3];
int n, i, j, k, flag = 0;
int  count = 0;
double sum1 = 0, sum2 = 0;
int main()
{
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        for (j = 0; j < 3; j++) {//由题知：列规定为3
            scanf("%lf", &a[i][j]);
        }
    }
    for (i = 0; i < n; i++) {
        for (k = i+1; k < n; k++) {
            for (j = 0; j < 3; j++) {
                double m;
                m = fabs(a[i][j] - a[k][j]);
                if (m <= 5) {//如果上下两行两人的分数绝对值之差小于5，则标志成真
                    flag = 1;
                    sum1 += a[i][j];
                    sum2 += a[k][j];
                }
                else {//如果不为真则跳出循环
                    flag = 0;
                    continue;
                }
            }
            if (flag == 1) {//建立在标志为真的基础上，进行总分条件的判断
                double p;
                p = fabs(sum1 - sum2);
                if (p < 10) {
                    count++;
                }
                else {
                    continue;
                }
            }
        }
    }
    printf("%d", count);
    return 0;
}