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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int, int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define itn int;
int dx[8] = { 1, 0, -1, 0, 1, -1, -1, 1 };
int dy[8] = { 0, 1, 0, -1, 1, 1, -1, -1 };
template <class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
struct io { io() { ios::sync_with_stdio(false); cin.tie(0); } };
const int INF = INT_MAX;
const ll LLINF = 1LL << 60;
const ll MOD = 1000000007;
const double EPS = 1e-9;
//#include <math.h>
signed main(void) 
{
    ll h, n, r;
    
    r = 0;
    ll sum = 0;
    cin >>h >>n;
    vll a(n);
    
    rep (i, n) {
        cin >>a[i];
        sum += a[i];
    }
    
    if (sum >= h) {
        cout <<"Yes";
        return 0;
    }
    cout <<"No";
    return 0;
}