Problem:
Score : 100 points

Problem Statement

An elementary school student Takahashi has come to a variety store.

He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?

Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Constraints

All input values are integers.

1 \leq A, B \leq 500

1 \leq C \leq 1000

Input

Input is given from Standard Input in the following format:

A B C

Output

If Takahashi can buy the toy, print Yes; if he cannot, print No.

Sample Input 1

50 100 120

Sample Output 1

Yes

He has 50 + 100 = 150 yen, so he can buy the 120-yen toy.

Sample Input 2

500 100 1000

Sample Output 2

No

He has 500 + 100 = 600 yen, but he cannot buy the 1000-yen toy.

Sample Input 3

19 123 143

Sample Output 3

No

There are 19-yen and 123-yen coins in Takahashi Kingdom, which are rather hard to use.

Sample Input 4

19 123 142

Sample Output 4

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
#include<unordered_map>
using namespace std;
#define MOD 1000000007
#define rep(i,m,n) for(int (i)=(int)(m);i<(int)(n);i++)
#define REP(i,n) rep(i,0,n)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
const long long INF = 1LL << 60;
typedef vector<ll> Array;
typedef vector<Array> Matrix;


//priority_queue<ll> max;//大きい順
//priority_queue<ll, Array, greater<ll>> min;//小さい順

/*firstについては昇順 secondについては降順
sort(all(wh), [&](P x, P y) {
	if (x.first == y.first)return x.second > y.second;
	return x.first < y.first;
});
*/

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}


//substr　文字列取り出し
//upper_bound ある値より大きい一番左のイテレータを返す、lowerは以上(setに対して使うとO(N)なので、setのメンバ関数を使う
//stoi


struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, ll from, ll to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}
ll max_flow_dfs(Graph & G, ll v, ll t, ll f, vector<bool> & used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph & G, ll s, ll t)//O(V(V+E))
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}
void BellmanFord(Graph& G, ll s, Array& d, Array &negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}
			}
		}
	}
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if (negative[i] == true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3)
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = INF;
		}
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			d[i][G[i][j].to] = G[i][j].cap;
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}
bool tsort(Graph& graph, vector<int>& order) {//トポロジカルソートO(E+V)
	int n = graph.size(), k = 0;
	Array in(n);
	for (auto& es : graph)
		for (auto& e : es)in[e.to]++;
	priority_queue<ll, Array, greater<ll>> que;
	REP(i, n)
		if (in[i] == 0)que.push(i);
	while (que.size()) {
		int v = que.top();
		que.pop();
		order.push_back(v);
		for (auto& e : graph[v])
			if (--in[e.to] == 0)que.push(e.to);
	}
	if (order.size() != n)return false;
	else return true;
}
class lca {
public:
	const int n = 0;
	const int log2_n = 0;
	std::vector<std::vector<int>> parent;
	std::vector<int> depth;

	lca() {}

	lca(const Graph& g, int root)
		: n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) {
		dfs(g, root, -1, 0);
		for (int k = 0; k + 1 < log2_n; k++) {
			for (int v = 0; v < (int)g.size(); v++) {
				if (parent[k][v] < 0)
					parent[k + 1][v] = -1;
				else
					parent[k + 1][v] = parent[k][parent[k][v]];
			}
		}
	}

	void dfs(const Graph& g, int v, int p, int d) {
		parent[0][v] = p;
		depth[v] = d;
		for (auto& e : g[v]) {
			if (e.to != p) dfs(g, e.to, v, d + 1);
		}
	}

	int get(int u, int v) {
		if (depth[u] > depth[v]) std::swap(u, v);
		for (int k = 0; k < log2_n; k++) {
			if ((depth[v] - depth[u]) >> k & 1) {
				v = parent[k][v];
			}
		}
		if (u == v) return u;
		for (int k = log2_n - 1; k >= 0; k--) {
			if (parent[k][u] != parent[k][v]) {
				u = parent[k][u];
				v = parent[k][v];
			}
		}
		return parent[0][u];
	}
};


class UnionFind {
	vector<int> data;
	ll num;
public:
	UnionFind(int size) : data(size, -1), num(size) { }
	bool unionSet(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		num -= (x != y);
		return x != y;
	}
	bool findSet(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	int numSet() {//集合の数を返す
		return num;
	}
};
class SumSegTree {
private:

	int _sum(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l)return 0;    // 交差しない
		if (a <= l && r <= b)return dat[k];   // a,l,r,bの順で完全に含まれる
		else {
			int s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
			int s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
			return s1 + s2;
		}
	}
public:
	int n, height;
	vector<int> dat;

	// 初期化（_nは最大要素数）
	SumSegTree(int _n) {
		n = 1;
		height = 1;
		while (n < _n) {
			n *= 2;
			height++;
		}
		dat = vector<int>(2 * n - 1, 0);
	}

	// 場所i(0-indexed)にxを足す
	void add(int i, int x) {
		i += n - 1; // i番目の葉ノードへ
		dat[i] += x;
		while (i > 0) { // 下から上がっていく
			i = (i - 1) / 2;
			dat[i] += x;
		}
	}

	// 区間[a,b)の総和。ノードk=[l,r)に着目している。
	int sum(int a, int b) {
		return _sum(a, b, 0, 0, n);
	}
};
class RmqTree {
private:

	ll _find(ll a, ll b, ll k, ll l, ll r) {
		if (r <= a || b <= l)return INF;    // 交差しない
		if (a <= l && r <= b)return dat[k];   // a,l,r,bの順で完全に含まれる
		else {
			ll s1 = _find(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
			ll s2 = _find(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
			return min(s1, s2);
		}
	}
public:
	ll n, height;
	vector<ll> dat;

	// 初期化（_nは最大要素数）
	RmqTree(ll _n) {
		n = 1;
		height = 1;
		while (n < _n) {
			n *= 2;
			height++;
		}
		dat = vector<ll>(2 * n - 1, INF);
	}

	// 場所i(0-indexed)をxにする
	void update(ll i, ll x) {
		i += n - 1; // i番目の葉ノードへ
		dat[i] = x;
		while (i > 0) { // 下から上がっていく
			i = (i - 1) / 2;
			dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
		}
	}

	// 区間[a,b)の最小値。ノードk=[l,r)に着目している。
	ll find(ll a, ll b) {
		return _find(a, b, 0, 0, n);
	}
};


//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());

}

vector<ll> lis_fast(const vector<ll>& a) {//最長部分増加列
	const ll n = a.size();
	vector<ll> A(n, INT_MAX);
	vector<ll> id(n);
	for (int i = 0; i < n; ++i) {
		id[i] = distance(A.begin(), lower_bound(A.begin(), A.end(), a[i]));
		A[id[i]] = a[i];
	}
	ll m = *max_element(id.begin(), id.end());
	vector<ll> b(m + 1);
	for (int i = n - 1; i >= 0; --i)
		if (id[i] == m) b[m--] = a[i];
	return b;
}


ll ModPow(ll x, ll n) {
	ll res = 1LL;
	while (n > 0) {
		if (n & 1) res = res * x % MOD;
		x = x * x % MOD;
		n >>= 1;
	}
	return res;
}
//nCrとか
class Combination {
public:
	Array fact;
	Array inv;
	ll mod;
	ll mod_inv(ll x) {
		ll n = mod - 2LL;
		ll res = 1LL;
		while (n > 0) {
			if (n & 1) res = res * x % mod;
			x = x * x % mod;
			n >>= 1;
		}
		return res;
	}
	ll nCr(ll n, ll r) {
		return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll n, ll _mod) {
		mod = _mod;
		fact.resize(n + 1);
		fact[0] = 1;
		REP(i, n) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		inv.resize(n + 1);
		REP(i, n + 1) {
			inv[i] = mod_inv(fact[i]);
		}
	}
};

ll gcd(ll m, ll n) {
	if (n == 0)return m;
	return gcd(n, m % n);
}//gcd

ll lcm(ll m, ll n) {
	return m / gcd(m, n) * n;
}


Matrix mIdentity(ll n) {
	Matrix A(n, Array(n));
	for (int i = 0; i < n; ++i) A[i][i] = 1;
	return A;
}

Matrix mMul(const Matrix& A, const Matrix& B) {
	Matrix C(A.size(), Array(B[0].size()));
	for (int i = 0; i < C.size(); ++i)
		for (int j = 0; j < C[i].size(); ++j)
			for (int k = 0; k < A[i].size(); ++k)
				(C[i][j] += (A[i][k] % MOD) * (B[k][j] % MOD)) %= MOD;
	return C;
}
// O( n^3 log e )
Matrix mPow(const Matrix & A, ll e) {
	return e == 0 ? mIdentity(A.size()) :
		e % 2 == 0 ? mPow(mMul(A, A), e / 2) : mMul(A, mPow(A, e - 1));
}


int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll a, b;
	cin >> a >> b;
	ll c;
	cin >> c;
	REP(i, c / a + 1) {
		REP(j, c / b + 1) {

			if (i*a + j * b == c) {
				cout << "Yes";
				return 0;
			}
		}

	}
	cout << "No";

	return 0;
}