#include <iostream>
#include <vector>
using namespace std;
// Function to check if it's possible to pay exact salaries using available banknotes
bool canPaySalaries(vector< int > &salaries, vector< int > &banknotes, int person, vector< bool > &used) {
if ( person == salaries.size( ) )
return true; // All salaries paid
int salary = salaries[person];
// Try each banknote combination that sums to the current salary
for ( int i = 0 ; i < banknotes.size( ) ; i++ ) {
if ( !used[i] && banknotes[i] <= salary ) {
if ( banknotes[i] == salary ) {
used[i] = true;
if ( canPaySalaries( salaries , banknotes , person + 1 , used ) )
return true;
used[i] = false;
} else {
// Try to combine with other banknotes
used[i] = true;
if ( canPaySalaries( salaries , banknotes , person , used ) )
return true;
used[i] = false;
}
}
}
return false;
}
int main() {
int N, M;
cin >> N >> M;
vector< int > salaries( N );
for ( int i = 0 ; i < N ; i++ ) {
cin >> salaries[i];
}
vector< int > banknotes( M );
for ( int i = 0 ; i < M ; i++ ) {
cin >> banknotes[i];
}
// If sum of banknotes is less than sum of salaries, it's impossible
long long sum_salaries = 0, sum_banknotes = 0;
for ( int s: salaries )
sum_salaries += s;
for ( int b: banknotes )
sum_banknotes += b;
if ( sum_banknotes < sum_salaries ) {
cout << "NO" << endl;
return 0;
}
// For N=1 case (Subtask 1), check if any combination of banknotes sums to salary
if ( N == 1 ) {
int target = salaries[0];
vector< bool > dp( target + 1 , false );
dp[0] = true;
for ( int b: banknotes ) {
for ( int j = target ; j >= b ; j-- ) {
dp[j] = dp[j] | dp[j - b];
}
}
cout << (dp[target] ? "YES" : "NO") << endl;
return 0;
}
// For general case, use backtracking
vector< bool > used( M , false );
cout << (canPaySalaries( salaries , banknotes , 0 , used ) ? "YES" : "NO") << endl;
return 0;
}
