B. Berland Bingo----模拟

B. Berland Bingo

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Lately, a national version of a bingo game has become very popular in Berland. There are n players playing the game, each player has a card with numbers. The numbers on each card are distinct, but distinct cards can have equal numbers. The card of the i-th player contains mi numbers.

During the game the host takes numbered balls one by one from a bag. He reads the number aloud in a high and clear voice and then puts the ball away. All participants cross out the number if it occurs on their cards. The person who crosses out all numbers from his card first, wins. If multiple people cross out all numbers from their cards at the same time, there are no winners in the game. At the beginning of the game the bag contains 100 balls numbered 1 through 100, the numbers of all balls are distinct.

You are given the cards for each player. Write a program that determines whether a player can win the game at the most favorable for him scenario or not.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of the players. Then follow n lines, each line describes a player's card. The line that describes a card starts from integer mi (1 ≤ mi ≤ 100) that shows how many numbers the i-th player's card has. Then follows a sequence of integers ai, 1, ai, 2, ..., ai, mi (1 ≤ ai, k ≤ 100) — the numbers on the i-th player's card. The numbers in the lines are separated by single spaces.

It is guaranteed that all the numbers on each card are distinct.

Output

Print n lines, the i-th line must contain word "YES" (without the quotes), if the i-th player can win, and "NO" (without the quotes) otherwise.

Examples

input

31 13 2 4 12 10 11

output

YESNOYES

input

21 11 1

output

NONO

题目链接：http://codeforces.com/contest/370/problem/B

我的英文真是差到一定程度了。。。看了半天没看懂，搜了题意。。

有n给人，每个人有m[i]张点数均不同的卡片，现在主持人没喊一个数持对应点数卡片的人能走卡片的步数，某人先走完自己手中全部卡片的获胜，问你在每个人都最有利的情况下每个人的获胜情况。

就是对每个人去其他人的卡片堆中找是否有自己卡片点数的子串，如果有自己就必输。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <set>using namespace std;const int MAX=105;int n,a[MAX],v;set<int>s[MAX];set<int>::iterator it;int main(){    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%d",&a[i]);        for(int j=0;j<a[i];j++){            scanf("%d",&v);            s[i].insert(v);        }    }    for(int i=0;i<n;i++){        bool flag=true;        for(int j=0;j<n;j++){            if(i==j) continue;            bool ok=false;            for(it=s[j].begin();it!=s[j].end();it++){                if(s[i].find(*it)==s[i].end()){                    ok=true;                    break;                }            }            if(ok==false){                flag=false;                break;            }        }        if(flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}