Problem:
Score : 200 points

Problem Statement

Takahashi has N days of summer vacation.

His teacher gave him M summer assignments. It will take A_i days for him to do the i-th assignment.

He cannot do multiple assignments on the same day, or hang out on a day he does an assignment.

What is the maximum number of days Takahashi can hang out during the vacation if he finishes all the assignments during this vacation?

If Takahashi cannot finish all the assignments during the vacation, print -1 instead.

Constraints

1 \leq N \leq 10^6

1 \leq M \leq 10^4

1 \leq A_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 ... A_M

Output

Print the maximum number of days Takahashi can hang out during the vacation, or -1.

Sample Input 1

41 2
5 6

Sample Output 1

30

For example, he can do the first assignment on the first 5 days, hang out on the next 30 days, and do the second assignment on the last 6 days of the vacation. In this way, he can safely spend 30 days hanging out.

Sample Input 2

10 2
5 6

Sample Output 2

-1

He cannot finish his assignments.

Sample Input 3

11 2
5 6

Sample Output 3

0

He can finish his assignments, but he will have no time to hang out.

Sample Input 4

314 15
9 26 5 35 8 9 79 3 23 8 46 2 6 43 3

Sample Output 4

9

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#define ll long long
#define pb push_back
#define debug(x) cout << '>' << #x << ':' << x << endl;
#define rep(i,n) for(ll i=0;i<(n);i++)
#define fr(i,a,b) for(ll i=(a);i<(b);i++)
#define frd(i,a,b) for(ll i=(a);i>=(b);i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define endl "\n"
#define inf ((long long) 1e18 + 10)
#define ve vector<ll>
using namespace std;
vector<bool> prime(100001,true);
void sieve(){
	ll n = 100000;
	for(ll i = 2; i*i <= n;i++ ){
		if(prime[i] == true){
			for(ll j = i * i ; j<=n ; j += i){
				prime[j] = false;
			}
		}
	}
	return ;
}

int main() {
    std::ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
	int n,k;
    cin>>k>>n;
    ve a(n);
    int val = 0;
    int s = 0;
    rep(i,n){
        cin>>val;
        s += val;
    }
    if(s > k){
        cout<<-1<<endl;
        return 0;
    }
    cout<<k - s<<endl;
    return 0;
}