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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <chrono> 
 
using namespace std;
using namespace std::chrono; 
 
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #pragma optimization_level 3
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define f0r(a, b) for (long long a = 0; a < (b); ++a)
#define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
#define f0rd(a, b) for (long long a = (b); a >= 0; --a)
#define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define pb push_back
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define fix(prec) {cout << setprecision(prec) << fixed;}
#define mp make_pair
#define f first
#define s second
#define presum(p, a, n) {p[0] = a[0]; for (int i = 1; i < (n); i++) p[i] = a[i] + p[i-1];}
#define all(v) v.begin(), v.end()
#define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
#define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
#define ai(a, n) for (int ele = 0; ele < n; ele++) cin >> a[ele];
#define ain(a, lb, rb) for (int ele = lb; ele <= rb; ele++) cin >> a[ele];
#define ao(a, n) {for (int ele = 0; ele < (n); ele++) { if (ele) cout << " "; cout << a[ele]; } cout << '\n';}
#define aout(a, lb, rb) {for (int ele = (lb); ele <= (rb); ele++) { if (ele > (lb)) cout << " "; cout << a[ele]; } cout << '\n';}
typedef long long ll;
typedef double ld;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
 
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
  cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
 
mt19937 rng(steady_clock::now().time_since_epoch().count());
// mt19937 rng(61378913);
/* usage - just do rng() */
 
void usaco(string filename) {
  // #pragma message("be careful, freopen may be wrong")
	freopen((filename + ".in").c_str(), "r", stdin);
	freopen((filename + ".out").c_str(), "w", stdout);
}
 
const lld pi = 3.14159265358979323846;
const ll mod = 1000000007;
// const ll mod = 998244353;
// ll mod;

ll n, m;

void solve(int tc) {
  cin >> m >> n;
  ll x = 0;
  f0r(i, n) {
    ll y; cin >> y; x += y;
  }
  cout << (x >= m ? "Yes\n" : "No\n");
}
 
int main() {
  send help
 
  // usaco("file");

  
 
  int tc;
  tc = 1;
  // cin >> tc;
  for (int t = 0; t < tc; t++) solve(t);
}