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: Accepted)

// ====================================================================================================================
// C++ includes used for precompiling -*- C++ -*-

// Copyright (C) 2003-2013 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 3, or (at your option)
// any later version.

// This library is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// Under Section 7 of GPL version 3, you are granted additional
// permissions described in the GCC Runtime Library Exception, version
// 3.1, as published by the Free Software Foundation.

// You should have received a copy of the GNU General Public License and
// a copy of the GCC Runtime Library Exception along with this program;
// see the files COPYING3 and COPYING.RUNTIME respectively.  If not, see
// <http://www.gnu.org/licenses/>.

/** @file stdc++.h
 *  This is an implementation file for a precompiled header.
 */

 // 17.4.1.2 Headers

 // C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif

// ここより↑は触らない(#include <bits/stdc++.h>の部分)
// ====================================================================================================================

#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;

typedef long long ll;

////==================================================================================
//// 素数を法とした整数の演算用
////==================================================================================
//const int mod = 1000000007;
////const int mod = 998244353;
//struct mint {
//	ll x; // typedef long long ll;
//	mint(ll x = 0) :x((x%mod + mod) % mod) {}
//	mint operator-() const { return mint(-x); }
//	mint& operator+=(const mint a) {
//		if ((x += a.x) >= mod) x -= mod;
//		return *this;
//	}
//	mint& operator-=(const mint a) {
//		if ((x += mod - a.x) >= mod) x -= mod;
//		return *this;
//	}
//	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
//	mint operator+(const mint a) const { return mint(*this) += a; }
//	mint operator-(const mint a) const { return mint(*this) -= a; }
//	mint operator*(const mint a) const { return mint(*this) *= a; }
//	mint pow(ll t) const {
//		if (!t) return 1;
//		mint a = pow(t >> 1);
//		a *= a;
//		if (t & 1) a *= *this;
//		return a;
//	}
//
//	// for prime mod
//	mint inv() const { return pow(mod - 2); }
//	mint& operator/=(const mint a) { return *this *= a.inv(); }
//	mint operator/(const mint a) const { return mint(*this) /= a; }
//};
//istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
//ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
////==================================================================================
//
////==================================================================================
//// コンビネーション
////==================================================================================
//struct combination {
//	vector<mint> fact, ifact;
//	combination(int n) :fact(n + 1), ifact(n + 1) {
//		assert(n < mod);
//		fact[0] = 1;
//		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
//		ifact[n] = fact[n].inv();
//		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
//	}
//	mint operator()(int n, int k) {
//		if (k < 0 || k > n) return 0;
//		return fact[n] * ifact[k] * ifact[n - k];
//	}
//};
////==================================================================================

int main(void) {
	int H, N;

	cin >> H >> N;

	int sum = 0;
	rep(i, N) {
		int wk;
		cin >> wk;
		sum += wk;
	}

	cout << (H > sum ? "No" : "Yes") << endl;

	return(0);
}