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)

/* ****revpro**** */
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
#define ll          long long
#define pb          push_back
#define endl     '\n'
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)    
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define mii      map<ll,ll>
using namespace std;

#define TIME        cerr << "\nTime elapsed: " << setprecision(5) <<1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define N  100005
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;  
#define ordered_set tree<double, null_type,less<double>, rb_tree_tag,tree_order_statistics_node_update> 
void solve()
{
  ll n,m;
  ll s=0;
  cin>>n>>m;
  rep(i,0,m){
    ll x;
    cin>>x;
    s+=x;
  } 
  if(s<=n)
  {
    cout<<n-s<<endl;
  }
  else cout<<-1<<endl;
}
signed main()
{
  
    /***********************************/
    unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
    mt19937_64 generator (seed);
    /***********************************/
 
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS=1;
//  cin>>TESTS;
  while(TESTS--)
  {
    solve();
    
  }
  TIME
  return 0;
}