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"
using namespace std;
using ll=long long;
#define print(n) cout<<n<<endl
#define rep(i,a,n) for(int i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
#define P pair<ll,ll>
#define vl vector<ll>
#define vi vector<int>
#define vvl vector<vector<ll>>
#define vvi vector<vector<int>>
const int M=200111;
const int inf=1000000007;
const int mod=1000000007;
const long long INF=1e18;
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
int ddx[8]={1,-1,0,0,1,-1,-1,1},ddy[8]={0,0,1,-1,1,-1,1,-1};


struct UnionFind {
	vector<int> par;
	UnionFind(int n) : par(n, -1) { }

	int root(int x) {
		if (par[x] < 0) return x;
		else return par[x] = root(par[x]);
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}

	bool merge(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (par[x] > par[y]) swap(x, y); // merge technique
		par[x] += par[y];
		par[y] = x;
		return true;
	}

	int size(int x) {
		return -par[root(x)];
	}
};

vvl comb(int n) {
	vvl v(n + 1,vl(n + 1, 0));
	for (int i = 0; i < v.size(); i++) {
		v[i][0] = 1;
		v[i][i] = 1;
	}
	for (int j = 1; j < v.size(); j++) {
		for (int k = 1; k < j; k++) {
			v[j][k] = (v[j - 1][k - 1] + v[j - 1][k])%mod;
		}
	}
	return v;
}

void sosu(ll x,bool f[]){
	ll y=x;
	for(ll i=2;i<=sqrt(y);i++){
		if(x%i==0){
			x/=i;
			f[i]=true;
			i--;
		}
	}
}

ll __gcd(ll x,ll y){
	if(x<y)swap(x,y);
	while(1){
		ll tmp=x%y;
		x=y;
		y=tmp;
		if(y==0)break;
	}
	return x;
}

int keta(ll a){
	int ke=1;
	while(a/=10)ke++;
	return ke;
}

vector<long long> enum_divisors(long long N) {
	vector<long long> res;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			res.push_back(i);
			// 重複しないならば i の相方である N/i も push
			if (N/i != i) res.push_back(N/i);
		}
	}
	// 小さい順に並び替える
	sort(res.begin(), res.end());
	return res;
}

ll zyou(ll x,ll n){
	ll y=1;
	rep(i,0,n){
		y*=x;
	}
	return y;
}
const double PI = 3.1415926535897932384626433;

int main() {
	int n,m,a[10001];
	cin>>n>>m;
	rep(i,0,m)cin>>a[i];
	int cnt=0;
	rep(i,0,m)cnt+=a[i];
	if(cnt>n)print(-1);
	else print(n-cnt);

}