Problem:
Score : 100 points

Problem Statement

Two students of AtCoder Kindergarten are fighting over candy packs.

There are three candy packs, each of which contains a, b, and c candies, respectively.

Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.

Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Constraints

1 ≦ a, b, c ≦ 100

Input

The input is given from Standard Input in the following format:

a b c

Output

If it is possible to distribute the packs so that each student gets the same number of candies, print Yes. Otherwise, print No.

Sample Input 1

10 30 20

Sample Output 1

Yes

Give the pack with 30 candies to one student, and give the two packs with 10 and 20 candies to the other. Then, each gets 30 candies.

Sample Input 2

30 30 100

Sample Output 2

No

In this case, the student who gets the pack with 100 candies always has more candies than the other.

Note that every pack must be given to one of them.

Sample Input 3

56 25 31

Sample Output 3

Yes

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#define SORT(v) sort(v.begin(), v.end())
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <cstdlib>
#include <cstdarg>
#include <cstdio>
#include <cmath>
#include <numeric>
#include <utility>

// #include "ane.cpp"

#define INF (int)1e9
#define INFLL (long long)1e18
#define NMAX 100005
#define MMAX 100005
#define MOD 100000
using namespace std;

// コメントアウトするとdebug()を実行しない
#define DEBUG





//
//  ライブラリ
//

// frequent used aliases
typedef long long ll;
typedef pair<int, int> p;
typedef pair<ll, int> lp;
typedef pair<ll, ll> llp;
typedef vector<int> vec;
typedef vector<vec> mat;

// frequent used constants
static const int di[] = {-1, 0, 1, -1, 1, -1, 0, 1};
static const int dj[] = {-1, -1, -1, 0, 0, 1, 1, 1};

// デバッグ用printf
void debug(const char* format, ...){
    #ifndef DEBUG
    return;
    #endif
    va_list arg;
    va_start(arg, format);
    vprintf(format, arg);  // コンソールに出力
    va_end(arg);
}

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

// Union-Find Tree
class UnionFindTree{
    struct node{
        int par;
    };
    std::vector<node> T;

    public:
    void init(int num){
        T.resize(num+1);    // ignore T[0]
        for (int i = 1; i <= num; ++i)
        {
            T[i].par = i;
        }
    }
    void unite(int x, int y){
        T[find(y)].par = find(x);
    }
    int find(int x){
        if (T[x].par == x) return x;
        else return T[x].par = find(T[x].par);
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
};

// Segment Tree for Range Minimum Query
    // **********************************************************
    // *** important: all functions' variable,                ***
    // *** such as "index", "l", "r", etc., must be 0-origin. ***
    // **********************************************************
    // ********************************************
    // *** important: NMAX must be power of 2. ***
    // ********************************************
template<typename T>
class SegmentTree{
    private:
    ll N;
    T INF_VAL;
    T dat[NMAX * 2]; // 0 origin, A[i] = dat[i + N]

    T _query(ll l, ll r, ll l_resp, ll r_resp, ll i_dat){
        debug("query(%lld, %lld, %lld, %lld, %lld) called\n",
            l, r, l_resp, r_resp, i_dat);
        if (r < l_resp || r_resp < l) return INF_VAL;
        else if(l <= l_resp && r_resp <= r) return dat[i_dat];
        else return min(_query(l ,r, l_resp, (l_resp + r_resp) / 2, i_dat * 2),
                        _query(l, r, (l_resp + r_resp) / 2 + 1, r_resp, i_dat * 2 + 1));
    }

    public:
    void init(ll _N, T _inf_val){
        N = 1; while(N < _N) N *= 2;
        INF_VAL = _inf_val;
        Fill(dat, _inf_val);
    }
    T get(ll index){
        return dat[index + N];
    }
    void set(int index, T val){
        int i_dat = index + N;
        dat[i_dat] = val;
        for (i_dat /= 2; i_dat > 0; i_dat /= 2)
        {
            dat[i_dat] = min(dat[i_dat * 2], dat[i_dat * 2 + 1]);
        }
    }
    T query(ll l, ll r){
        return _query(l, r, 0, N - 1, 1);
    }
    void dump(){
        cout << "*** SegTree dump begin ***\n";
        cout << "N = " << N << ", INF_VAL = " << INF_VAL << endl;
        for (int i = 1; i < N * 2; i *= 2)
        {
            for (int j = i; j < i * 2; ++j)
            {
                if(dat[j] == INF_VAL) cout << "INF ";
                else cout << dat[j] << " ";
            }
            cout << endl;
        }
        cout << "*** SegTree dump end ***\n";
    }
};

// Binary Indexed Tree for Range Sum Query
    // *******************************************
    // *** important: all functions' variable, ***
    // *** such as "i", must be 1-origin.      ***
    // *******************************************
template<typename T>
class BinaryIndexedTree{
    private:
    ll N;
    T dat[NMAX + 1]; // 1 origin, A[i] = sum(i) - sum(i-1)

    public:
    void init(ll _N){
        N = _N;
        Fill(dat, 0);
    }
    void add(int i, T val){
        while(i <= N) {
            dat[i] += val;
            i += i & -i;
        }
    }
    T sum(ll i){
        T ret = 0;
        while(i > 0) {
            ret += dat[i];
            i -= i & -i; // set last HIGH bit to LOW
        }
        return ret;
    }
    T sum(ll left, ll right){
        return sum(right) - sum(left - 1);
    }
    void dump(){
        cout << "*** BITree dump begin ***\n";
        cout << "N = " << N << endl;
        for (int i = 1; i <= N; i *= 2)
        {
            cout << dat[i] << " ";
        }
        cout << "*** BITree dump end ***\n";
    }
};

//
// ライブラリ終了
//







ll N,M,K,A,B,C,D,E;
int dp[NMAX] = {};

ll ans = 0;

void solve(){
    // 解答アルゴリズム
}
void debug(){
    // デバッグ用出力
}
void answer(){
    // 解答出力
    if (N + K == M || N + M == K || M + K == N)
    {
            cout << "Yes\n";
    }else{
        cout << "No\n";
    }
}
int main(int argc, char const *argv[])
{
    // 入力の読み込み，番兵法

    // Fill(dp, -1);
    scanf("%lld%lld%lld", &N,&M, &K);
    solve();
    #ifdef DEBUG
    debug();
    #endif
    answer();

    return 0;
}