Problem:
Score : 200 points

Problem Statement

Raccoon is fighting with a monster.

The health of the monster is H.

Raccoon can use N kinds of special moves. Using the i-th move decreases the monster's health by A_i.
There is no other way to decrease the monster's health.

Raccoon wins when the monster's health becomes 0 or below.

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Constraints

1 \leq H \leq 10^9

1 \leq N \leq 10^5

1 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H N
A_1 A_2 ... A_N

Output

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Sample Input 1

10 3
4 5 6

Sample Output 1

Yes

The monster's health will become 0 or below after, for example, using the second and third moves.

Sample Input 2

20 3
4 5 6

Sample Output 2

No

Sample Input 3

210 5
31 41 59 26 53

Sample Output 3

Yes

Sample Input 4

211 5
31 41 59 26 53

Sample Output 4

No

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

#include "bits/stdc++.h"
#include <chrono> 
 
using namespace std;
using namespace std::chrono; 
 
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
 
#define gcd(a,b) __gcd(a,b)
#define vi vector<int>
#define vll vector<ll>
#define vs vector<string>
#define vc vector<char>
#define pb(n) push_back(n)
#define pp pop_back()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define type(a) typeid(a).name();
#define print(arr) cout << "[" ;for(unsigned int i=0;i<(unsigned int) arr.size();i++){ (i==arr.size()-1)?cout << arr[i]:cout << arr[i] << ",";}cout << "]" << endl
#define Arr(arr,l) cout << "[" ;for(unsigned int i=0;i<(unsigned int) l;i++){ (i==l-1)?cout << arr[i]:cout << arr[i] << ",";}cout << "]" << endl
#define console(arr) cout << "["<<endl; for(ll i=0;i<(ll)arr.size();++i){cout << " [";for(ll j=0;j<(ll)arr[i].size();++j){(j==(int)arr[i].size()-1)?cout << arr[i][j]:cout << arr[i][j] << ",";}cout << "]"<<endl;} cout << "]"<<endl
#define starting() auto start = high_resolution_clock::now()
#define ending() auto stop = high_resolution_clock::now();auto duration = duration_cast<microseconds>(stop - start);cout << endl << endl << "Execution Time : "<<duration.count()/1000 << " ms" << endl
 
template <typename T1,typename T2>
void mapped(map<T1,T2>m){for(auto ite=m.begin();ite!=m.end();++ite){cout << "'" << ite->first << "'"<<" : "<<ite->second <<endl;}}
 
template <typename T1 , typename T2>
void mappedArr(map<T1,vector<T2>>m){for(auto ity=m.begin();ity!=m.end();++ity){cout << "'"<< ity->first <<"'"<<" : "<<"[";
for(int i=0;i<(int)ity->second.size();++i){(i==(int)ity->second.size()-1)?cout << ity->second[i]:cout << ity->second[i] << ",";}cout << "]"<<endl;}}
 
string int_to_string(int x){ stringstream ss; ss << x; string ni = ss.str(); return ni;}
int char_to_int(char c){ int n = (int)c - 48; return n; }
int string_to_int(string x){ int n; stringstream s(x); s >> n; return n; }
 
template <typename T1,typename T2>
void ArrPair(vector<pair<T1,T2>>vp){for(int i=0;i<(int)vp.size();++i){
cout << "(" << vp[i].first << "," << vp[i].second << ")"<<endl ;}}
 
const int MOD = 1e9 + 7;
const ll INF = 1e18L + 5;
 
unordered_map<int,int> um;
unordered_set<int> us;
set<int> s;
// v.front() gives the first element of vector 
// v.back()  gives the last element of vector
// insert erase substr
//g++ -o Legendary-Grandmaster Legendary-Grandmaster.cpp&Legendary-Grandmaster.exe 
//cd C:\Users\158si\Desktop\Algorithim\codeForces\elite
// for-bash cd C:/Users/158si/Desktop/Algorithim/codeForces/elite

int main(){
	int64_t k,n;
	cin >> k >> n;
	int64_t summed = 0;
	for(int i=0;i<n;++i){
		int64_t x;
		cin >> x;
		summed += x; 
	}
	if(summed>=k){
		cout << "YES";
	}else{
	   cout << "NO";
     }
	return 0;
}