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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
import java.io.PrintWriter

val pw = PrintWriter(System.out)
val MOD = 1000000007L
val INF = 2147483647
val LINF = 9223372036854775807L

fun main(args: Array<String>){
    solve()
    pw.flush()    
}

fun solve(){
    val h = nextIntList()
    val a = nextIntList()
    println(if(a.sum() >= h.first()) "Yes" else "No")


}


// Print
fun println(v: String){
    pw.println(v)
}
fun print(v: String){
    pw.print(v)
}

// Read
fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ")
fun nextIntList() = next().split(" ").map{ it.toInt() }
fun nextLongList() = next().split(" ").map{ it.toLong() }
fun nextDoubleList() = next().split(" ").map{ it.toDouble() }
fun nextAryln(n: Int) = Array(n){ next() }
fun nextIntAryln(n: Int) = IntArray(n){ nextInt() }
fun nextLongAryln(n: Int) = LongArray(n){ nextLong() }
fun nextDoubleAryln(n: Int) = DoubleArray(n) { nextDouble() }

// Math
fun abs(n: Long) : Long = Math.abs(n)
fun abs(n: Double) : Double = Math.abs(n)
fun max(a: Long = -LINF, b: Long = -LINF, c: Long = -LINF, d: Long = -LINF, e: Long = -LINF): Long = listOf(a, b, c, d, e).max()!!.toLong()
fun min(a: Long = LINF, b: Long = LINF, c: Long = LINF, d: Long = LINF, e: Long = LINF): Long = listOf(a, b, c, d, e).min()!!.toLong()
fun prime(from: Long, to: Long = from) : List<Long>{
    return (from..to).filter{ i ->
        val max = Math.sqrt(i.toDouble()).toLong()
        (2..max).all{ j -> i % j != 0L}
    }
}
fun gcd(a: Long, b: Long) : Long = if(a % b == 0L) b else gcd(b, (a % b))
fun lcm(a: Long, b: Long) : Long = a / gcd(a, b) * b
fun modpow(a: Long, n: Long, p:Long = MOD) : Long {
    var res = 1L
    var ar = a
    var nr = n
    while(nr > 0){
        if((nr and 1) == 1L) res = res * ar % p
        ar = ar * ar % p
        nr = nr shr 1
    }
    return res
}
fun modinv(a: Long, p: Long = MOD) : Long = modpow(a, p - 2, p)
fun ncr(n: Long, r: Long) : Long {
    var a = 1L
    var b = 1L
    for (i in 1..r) {
        a = a * (n + 1 - i) % MOD
        b = b * i % MOD
    }
    return modinv(b, MOD) * a % MOD
}