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<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<iomanip>
#include<queue>
#include<stack>
#include<time.h>
#define rep(i,n)for(int i=0;i<n;i++)
#define int long long
#define ggr getchar();getchar();return 0;
#define prique priority_queue
constexpr auto mod = 1000000007;
#define inf 1e15
#define key 1e9
using namespace std;
typedef pair<int, int>P;
void yes() { cout << "Yay!" << endl; }
void no() { cout << ":(" << endl; }
template<class T> inline void chmax(T& a, T b) {
	a = std::max(a, b);
}
template<class T> inline void chmin(T& a, T b) {
	a = std::min(a, b);
}

//combination(Nが小さい時はこれを使う)
const int MAX = 300000;
int fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod % i] * (mod / i) % mod;
		finv[i] = finv[i - 1] * inv[i] % mod;
	}
}
int COMB(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}

bool prime(int n) {
	int cnt = 0;
	for (int i = 1; i <= sqrt(n); i++) {
		if (n % i == 0)cnt++;
	}
	if (cnt != 1)return false;
	else return n != 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x / gcd(x, y) * y;
}

//繰り返し二乗法（Nが大きい時の場合のcombination)
int mod_pow(int x, int y, int m) {
	int res = 1;
	while (y) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}
int kai(int x, int y) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= (i % mod); res %= mod;
	}
	return res;
}
int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x, y) * mod_pow(kai(y, y), mod - 2, mod) % mod;
}
//UnionFind
class UnionFind {
protected:
	int* par, * rank, * size;
public:
	UnionFind(unsigned int size) {
		par = new int[size];
		rank = new int[size];
		this->size = new int[size];
		rep(i, size) {
			par[i] = i;
			rank[i] = 0;
			this->size[i] = 1;
		}
	}
	int find(int n) {
		if (par[n] == n)return n;
		return par[n] = find(par[n]);
	}
	void unite(int n, int m) {
		n = find(n);
		m = find(m);
		if (n == m)return;
		if (rank[n] < rank[m]) {
			par[n] = m;
			size[m] += size[n];
		}
		else {
			par[m] = n;
			size[n] += size[m];
			if (rank[n] == rank[m])rank[n]++;
		}
	}
	bool same(int n, int m) {
		return find(n) == find(m);
	}
	int getsize(int n) {
		return size[find(n)];
	}
};
int dight_sum(int n) {
	int sum = 0;
	rep(i, 20)sum += (n % (int)pow(10, i + 1)) / (pow(10, i));
	return sum;
}
int a[11111];
signed main() {
	int n, m;
	cin >> n >> m;
	rep(i, m)cin >> a[i];
	int memo = 0;
	rep(i, m)memo += a[i];
	if (memo > n)puts("-1");
	else cout << n - memo << endl;
	ggr
}