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)

// Vaidik Patel(India)
// DA-IICT
#include <bits/stdc++.h>
using namespace std;
//********************************************
#include<ext/pb_ds/assoc_container.hpp> // include the associative containers or group of templates such as set, multimap, map etc.
#include<ext/pb_ds/tree_policy.hpp> // include the tree_order_statistics_node update
using namespace __gnu_pbds; //necessary for the GNU based Policy based data structures.
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//********************************************
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
int INF = (int)1e9;
ll INFINF = (ll)1e18 + 10;
const ld PI = 3.14159265358979323846;
#define precision(x,d) cout<<fixed<<setprecision(d)<<x
#define minheap priority_queue<ll,vector<ll>,greater<ll> >
#define maxheap priority_queue<ll,vector<ll>,less<ll> >
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define rFOR(i,a,b) for(ll i=a;i>=b;i--)
#define deb1(x) cout<<#x<<" : "<<x<<endl;
#define deb2(x,y) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<endl;
#define deb3(x,y,z) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<"\t"<<#z<<" : "<<z<<endl;
#define deb4(x,y,z,w) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<"\t"<<#z<<" : "<<z<<"\t"<<#w<<" : "<<w<<endl;
#define deb5(a, b, c, d, e)    cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl;
#define deb6(a, b, c, d, e, f) cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl;
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define checkbit(n,b) ( (n >> b) & 1)
#define mkp make_pair
#define ff first
#define ss second
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define TIME cerr << "Time Elapsed : "<<1.0*clock()/CLOCKS_PER_SEC<<"\n";
#define lc 2*idx
#define rc 2*idx+1
ll powermodm(ll x,ll n,ll mod)
{ll result=1;while(n>0){if(n % 2 ==1)result=(result * x)%mod;x=(x*x)%mod;n=n/2;}return result;}
ll GCD(ll , ll );
ll LCM(ll ,ll);
ll power(ll ,ll);
ll choose(ll,ll);
int ones(ll);
void extendedEuclid(ll, ll);
ll MMI(ll, ll);
void fastscan(int &);
bool checkithbit(ll,ll); // ith bit is zero or not
vector<ll>SegmentTree;
void Build_Segment_Tree(vector<ll>&a,ll idx,ll l,ll r);
ll sum(ll idx,ll sl,ll sr,ll l,ll r);
void update(ll idx,ll sl,ll sr,ll value,ll id);
vector<ll>prime;
void findprime(void);
ll phi(ll num);
vector<int>isprime;
void precomputeprime(int N);
vector<int>prephi;
void precomputephi(int N);
inline int read();
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
#define pb push_back
#define M  1000000000
//////////////////////////////////////////////////////////////////////////////

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    
    int t;
    //cin>>t;
    t=1;
    while(t--)
    {
        ll d,n;
        cin>>d>>n;
        ll tot=0;
        FOR(i,0,n)
        {
            ll x;
            cin>>x;
            tot+=x;
        }
        if(tot>d)
        {
            cout<<-1;
      }
      else
      {
        cout<<d-tot;
      }
    }
    return 0;
}
//////////////////////////////////////////////////////////////////////////
//precomputephi()
void precomputephi(int N)
{
    prephi.resize(N);

    FOR(i,0,N)
    {
        prephi[i]=i;
    }

    for(int i=1;i<N;i++)
    {
        if(isprime[i])
        {
            for(int j=i;j<N;j+=i)
            {
                prephi[j]=prephi[j]-(prephi[j]/i);
            }
        }
    }
}

//precomputedprime()
void precomputeprime(int N)
{
    isprime.resize(N);
    isprime.assign(N,1);
    isprime[0]=0;
    isprime[1]=0;
    for(int i=2;i*i<=500005;i++)
    {
        if(isprime[i])
        {
            for(int j=i*i;j<=500005;j+=i)
            {
                isprime[j]=0;
            }
        }
    }
}
//"""euler totient function""": counts the number of integers between 1 and n inclusive, which are coprime to n.
ll phi(ll num) // find using factorization in O(sqrt(N)).....
{
    ll result=num;
    for(ll i=2;i*i<=num;i++)
    {
        if(num%i==0)
        {
            while(num%i==0)
            {
                num/=i;
            }
            result-=(result/i);
        }
    }
    if(num>1)
    {
        result-=(result/num);
    }
    return result;
}

void findprime(void)
{
    int N=1e6+5;
    vector<bool>isprime(N,1); //up to 1e6+5 number
    for(int i=2;i*i<=N;i++)
    {
        if(isprime[i])
        {
            for(int j=i*i;j<=N;j+=i)
            {
                if(isprime[j])
                {
                    isprime[j]=0;
                }
            }
        }
    }

    for(int i=2;i<=N;i++)
    {
        if(isprime[i])
        {
            prime.pb(i);
        }
    }

}

bool checkithbit(ll a,ll i)
{
if(a & (1 << i))
{
return  true;
}
else
{
return false;
}
}
ll GCD(ll a , ll b)
{
    if(b==0)
        return a;
    else
        return GCD(b,a%b);
}
ll LCM(ll a,ll b){return (max(a,b)/GCD(a,b))*min(a,b);}

ll power(ll a , ll n)
{
    unsigned long long int result=1,x=a;
    while(n>0)
    {
        if(n % 2 ==1)
            result=result * x;
        x = x * x;
        n=n/2;
    }
    return result;
}
ll choose(ll n,ll k)
{
    if(k==0)  return 1;
    return (n* choose(n-1,k-1))/k;
}
int ones(ll n)
{
    int c=0;
    while(n)
    {
        n=n&(n-1);
        c++;
    }
    return c;
}
//store gcd of a,b in d
//store solution of a*x + b*y = d in x,y
// if gcd(a,b)=1 then x = a^(-1)(mod b) and y = b^(-1)(mod a)
ll d, x, y;
void extendedEuclid(ll A, ll B) {
    if(B == 0) {
        d = A;
        x = 1;
        y = 0;
    }
    else
    {
        extendedEuclid(B, A%B);
        int temp = x;
        x = y;
        y = temp - (A/B)*y;
    }
}
// return modulo inverse of a wrt modulo p(should be prime)
// retunt -1 if modulo inverse does not exist
ll MMI(ll a, ll p)
{
    extendedEuclid(a,p);
    if(d==1 && p!=1)
        return ((x%p)+p)%p;
    else
        return -1;
}

//Fastest way to read Input
void fastscan(int &number)
{
    bool negative = false;
    register int c;
    number = 0;
    c = getchar();
    if (c=='-')
    {
        negative = true;
        c = getchar();
    }
    for (; (c>47 && c<58); c=getchar())
        number = number *10 + c - 48;
    if (negative)
        number *= -1;
}
inline int read () {
    int ret = 0, t = 1;
    char c = getchar();
    while ((c < '0' || c > '9') && c != '-') c = getchar();
    if (c == '-') t = -1, c = getchar();
    while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
    return ret * t;
}
void Build_Segment_Tree(vector<ll>&a,ll idx,ll l,ll r) //where l and r is query range.
{
  if(l==r)
  {
    SegmentTree[idx]=a[l];
  }
  else
  {
    ll mid=(l+r)/2ll;
    Build_Segment_Tree(a,(2ll*idx),l,mid);
    Build_Segment_Tree(a,(2ll*idx)+1,mid+1,r);
    SegmentTree[idx]=SegmentTree[2ll*idx]+SegmentTree[2ll*idx+1];
  }
}
ll sum(ll idx,ll sl,ll sr,ll l,ll r) //here [sl...sr] segment range associate at idx index. where l and r is query range.
{
    if(l>r)
    {
        return 0;
    }
    if(sl==l && sr==r)
    {
        return SegmentTree[idx];
    }


    ll mid=(sl+sr)/2;
    return sum(2ll*idx,sl,mid,l,min(r,mid))+sum(2ll*idx+1,mid+1,sr,max(mid+1,l),r);
}

void update(ll idx,ll sl,ll sr,ll value,ll id)
{
    if(sl==sr && sl==id)
    {
        SegmentTree[idx]=value;
    }
    else
    {
        ll mid=(sl+sr)/2ll;
        if(id<=mid)
        {
            update(2ll*idx,sl,mid,value,id);
        }
        else
        {
            update(2ll*idx+1,mid+1,sr,value,id);
        }
        SegmentTree[idx]=SegmentTree[2ll*idx]+SegmentTree[2ll*idx+1];
    }
}
/*
note:
//a.resize(unique(a.begin(), a.end()) - a.begin()); use for create array unique element...
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); shuffle(g1.begin(), g1.end(), rng);(g1 is array name) for choose random number
//int grand(int x) {
    return uniform_int_distribution<int>(0, x - 1)(rng);
}
//use once(num) give no of 1 in its binary form
//use give (int)log2(num)+1 give no of digits in its binary form
//bitset<64>(num).to_string() convert decimal to num64 bit binary number
//stoull(str, nullptr, 2) convert str in binary form to decimal number
// "max" priority queue: priority_queue<ll, vector<ll>, less<ll> >pq;   #select max element as top()#
// "min" priority queue: priority_queue<ll, vector<ll>, greater<    ll> >pq; #select min element as top()#
//set<int> inbuit sort accending order (set<int,less<int>>)
//set<int,greater<int>> sort element in decreasing order
// int  __builtin_popcount(unsigned int)
--->It returns the numbers of set bits in an integer (the number of ones in the binary representation of the integer).
///for finding xnor from xor
        x1=Xor
    x2=Xnor
    ll x1=a ^ b;

        ll num=max(a,b);

        ll xx=log2(num)+ 1;   (length of num in binary form)
    ****ll t=(1 << xx) - 1;  (set all xx length bits to 1)
        ll x2=x1 ^ t;            (get xnor via xor using t)


--->Dilworth's theorem:-
        this theorem says that the minimum number of non-decreasing sequences we need to cover the
        whole sequence equals the length of least decreasing subsequence.
----->a + b = (a xor b) +( 2 * (a & b)); ///very imp
//~ while (clock()<=69*CLOCKS_PER_SEC)
//~ #pragma comment(linker, "/stack:200000000")
//~ #pragma GCC optimize("O3")
//~ #pragma GCC target ("avx2")
//~ #pragma GCC optimize("Ofast")
//~ #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//~ #pragma GCC optimize("unroll-loops")

x&(-x) which is equal to highest power of two that divides x.
x&(x-1) which is equal 0 only when x is a power of two(except x == 0)
###
string str;
vector<string>ip;
istringstream iss(str);
for(string s;iss>>s;)
{
 ip.push_back(s);
}
###
*/