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 <cstdio>
#include <vector>
#include <map>
#include <string>
#include <utility>
#include <algorithm>
#include <functional>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
using namespace std;
using ll=long long;
using intpair=pair<int,int>;
using intpv=vector<intpair>;
using llpair=pair<ll,ll>;
using llpv=vector<llpair>;
using intvec=vector<int>;
using llvec=vector<ll>;
using intq=queue<int>;
using llq=queue<ll>;
using intmat=vector<intvec>;
using llmat=vector<llvec>;
using pairmat=vector<llpv>;
#define matrix(T) vector<vector<T>>
#define PI 3.141592653589793
#define INTINF 1<<30
#define LLINF 1LL<<60
#define MPRIME 1000000007
#define pqueue priority_queue
#define pushb push_back
#define all(name) name.begin(),name.end()
#define rall(name) name.rbegin(),name.rend()
#define ABS(x) ( (x)>0 ? (x) : -(x) )
#define gsort(vbeg,vend) sort(vbeg,vend,greater<>())
#define init(v) for(auto &a: v) cin>>a
#define out(n) cout<<n<<endl

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}

template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

ll GCD(ll a, ll b) { //最大公約数
	if(a==0||b==0) return 0;
	if(a<b) swap(a,b);
	ll tmp = a%b;
	while(tmp!=0) {
		a = b;
		b = tmp;
		tmp = a%b;
	}
	return b;
}

ll binpow(ll a, ll ex, ll p) { //繰り返し二乗法
	ll result=1LL;
	while(ex>0) {
		if(ex&1) result=result*a%p;
		ex>>=1;
		a=a*a%p;
	}
	return result;
}

ll Fact(ll x, ll p) { //階乗
	ll f=1;
	for(ll i=2; i<=x; i++) {
		f*=i;
		f%=p;
	}
	return f;
}

ll nPr(ll n, ll r) {
	if(n<r) return 0;

	ll result=1LL;
	for(ll i=0; i<r; i++) result*=n-i;
	return result;
}

ll nPrP(ll n, ll r, ll p) { // mod pにおけるnPr
	if(n<r) return 0;

	ll result=1LL;
	for(int i=0; i<r; i++) {
		result*=n-i;
		result%=p;
	}
	return result;
}

ll nCr(ll n, ll r) {
	if (n == r) { return 1; }
	if (r > n) { return 0; }

	if (r > n / 2) { r = n - r; }

	if (n == 0) { return 0; }
	if (r == 0) { return 1; }
	if (r == 1) { return n; }

	double result = 1;
	for (double i = 1; i <= r; i++) {
		result *= (n - i + 1) / i;
	}

	return (ll)result;
}

llvec fact,inv,factinv;
void prenCrP(ll n, ll p) {
	fact.resize(n+1);
	inv.resize(n+1);
	factinv.resize(n+1);
	fact[0]=fact[1]=inv[1]=factinv[0]=factinv[1]=1LL;
	for(ll i=2LL; i<=n; i++) {
		fact[i]=fact[i-1]*i%p;
		inv[i]=p-inv[p%i]*(p/i)%p;
		factinv[i]=factinv[i-1]*inv[i]%p;
	}
}
ll nCrP(ll n, ll r, ll p) { // mod pにおけるnCr
	if (r > n) return 0;
	return fact[n]*factinv[r]%p*factinv[n-r]%p;
}

llvec fact2,inv2,factinv2;
void prenCrP2(ll n, ll r, ll p) { // nがバカでかいときに使う, 計算量はr依存
	fact2.resize(r+1);
	inv2.resize(r+1);
	factinv2.resize(r+1);
	fact2[0]=n%p, fact2[1]=n%p*(n-1)%p;
	inv2[1]=factinv2[0]=factinv2[1]=1LL;
	for(ll i=2LL; i<=r; i++) {
		fact2[i]=fact2[i-1]*(n-i)%p;
		inv2[i]=p-inv2[p%i]*(p/i)%p;
		factinv2[i]=factinv2[i-1]*inv2[i]%p;
	}
}
ll nCrP2(ll r, ll p) {
	return fact2[r-1]*factinv2[r]%p;
}

ll LowerBinarySearch(llvec array, ll key, ll max, ll min) { //にぶたんlower
	if(max>min) {
		return -1LL;
	} else {
		ll mid = max + (min-max)/2;
		if(array[mid]>key) {
			return LowerBinarySearch(array,key,max,mid-1);
		} if(array[mid]<key) {
			return LowerBinarySearch(array,key,mid+1,min);
		} else {
			return mid;
		}
	}
}

ll GreaterBinarySearch(llvec array, ll key, ll max, ll min) { //にぶたんgreater
	if(max>min) {
		return -1LL;
	} else {
		ll mid = max + (min-max)/2;
		if(array[mid]<key) {
			return GreaterBinarySearch(array,key,max,mid-1);
		} if(array[mid]>key) {
			return GreaterBinarySearch(array,key,mid+1,min);
		} else {
			return mid;
		}
	}
}

int DigitNum(ll n) { //桁数
	int digit=0;
	ll wari=1LL;
	while(n/wari) {
		digit++;
		wari*=10;
	}
	return digit;
}

bool IsPrime(ll num) { //素数判定
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

	double sqrtNum = sqrt(num);
	for (ll i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

vector<ll> Divisor(ll x) { // 約数列挙
	vector<ll> result;
	ll i=1LL;
	for( ; i*i<x; i++) {
		if(x%i) continue;
		result.push_back(i);
		result.push_back(x/i);
	}
	if(i*i==x&&x%i==0)
		result.push_back(i);
	
	sort(result.begin(),result.end());
	return result;
}

vector<llpair> PrimeFact(ll x) { // 素因数分解 {素因数,指数}
	vector<llpair> result;
	ll ex=0LL;
	if(x%2==0) {
		while(x%2==0) {
			x/=2;
			ex++;
		}
		result.push_back({2,ex});
	}

	for(ll i=3LL; i*i<=x; i+=2) {
		if(x%i) continue;

		ex=0LL;
		while(x%i==0) {
			x/=i;
			ex++;
		}
		result.push_back({i,ex});
	}

	if(x!=1) result.push_back({x,1});

	return result;
}

bool Palind(string s) { //回文判定
	return s == string(s.rbegin(), s.rend());
}

struct Union_Find{
	vector<int> parent; //親
	vector<int> num; //根としたときの木サイズ

	Union_Find(int N) : parent(N),num(N,1) {
		for(int i=0; i<N; i++) {
			parent[i]=i;
		}
	}

	int root(int x) {
		if(parent[x]==x) return x;
		return parent[x]=root(parent[x]);
	}

	void merge(int x, int y) {
		int xrt=root(x);
		int yrt=root(y);
		if(xrt==yrt) return;
		parent[xrt]=yrt;
		num[yrt]+=num[xrt];
	}

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

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


int N,M;

int main() {
	cin>>N>>M;
	for(int i=0; i<M; i++) {
		int A; cin>>A;
		N-=A;
	}
cout<<max(-1,N)<<endl;

}