CodeForce 890A ACM ICPC

Description：

In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.

After practice competition, participant number i got a score of a_i. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.

Input

The single line contains six integers a₁, ..., a₆ (0 ≤ a_i ≤ 1000) — scores of the participants

Output

Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

Example

Input

1 3 2 1 2 1

Output

YES

Input

1 1 1 1 1 99

Output

NO

Note

In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.

In the second sample, score of participant number 6 is too high: his team score will be definitely greater.

題目大意：

有六個人隨機組隊去參加ICPC， 判斷兩個隊伍的得分是否能夠相等， 如果能則輸出“YES”， 否則輸出“NO”。

解題思路：

判斷隨機的三個數的和是否與另外三個數的和相等， 首先能想到的就是判斷總分數是否等於某三個數的和的二倍，對於枚舉三個數直接暴力去跑即可。

代碼：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*
 *ios::sync_with_stdio(false);
 */

typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x7fffffff;
const int mod = 1000000;
const int Max = (int) 2e5 + 9;

int arr[10];

int main() {
    int temp, sum = 0, flag = 0;
    for (int i = 1; i <= 6; ++i) {
        scanf("%d", &arr[i]);
        sum += arr[i];
    }
    // enumeration two teams
    for (int i = 1; i <= 6; ++i) {
        for (int j = i + 1; j <= 6; ++j) {
            for (int k = j + 1; k <= 6; ++k) {
                temp = arr[i] + arr[j] + arr[k];
                // whether equal
                if (sum - temp == temp) {
                    flag = 1;
                    break;
                }
            }
            if (flag) break;
        }
        if (flag) break;
    }
    if (flag) printf("YES\n");
    else printf("NO\n");
    return 0;
}