CF 231A team

题意：有N行，3列的矩阵，

          行代表题目，数量，列代表三个人；

         1为true，0为false，如果同一个题true>=2，则可解决题目数量+1；

          大致思路就是按行搜索就行；纯模拟，之前没看清题意，以为是N X N 矩阵；

          其实是N X 3 矩阵；

A. Team

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.

This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.

Input

The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Thenn lines contain three integers each, each integer is either0 or 1. If the first number in the line equals1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.

Output

Print a single integer — the number of problems the friends will implement on the contest.

Examples

Input

31 1 01 1 11 0 0

Output

2

Input

21 0 00 1 1

Output

1

Note

In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.

In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

AC代码：

#include <bits/stdc++.h>using namespace std ;int dp[1100][1100];int main(){int n,i,j ;int sum = 0 ;int count ;cin>>n;for(int i = 0 ; i < n ;i++){count=0;for(j = 0 ; j<3;j++){cin>>dp[i][j];if(dp[i][j]){count++;}}if(count>=2) sum++;}cout<<sum<<endl;return 0 ;}