A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?

Input

The input contains two integers a and b (0 ≤ a, b ≤ 103), separated by a single space.

Output

Output the sum of the given integers.

Examples

Input

5 14


Output

19


Input

381 492


Output

873
#include <bits/stdc++.h>
using namespace std;
const long long Nfact = 200005;
const long long mod = 1000000007;
const long long Nsieve = 1000005;
vector<bool> is_prime(Nsieve, true);
void sieve() {
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i * i <= Nsieve; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= Nsieve; j += i) is_prime[j] = false;
    }
  }
}
long long bSearch(long long l, long long r, long long x,
                  vector<long long> arr) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
void primeFactors(long long n) {
  while (n % 2 == 0) {
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      n = n / i;
    }
  }
  if (n > 2) cout << n << " ";
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modulo(long long a, long long b) {
  long long c = a % b;
  return (c < 0) ? c + b : c;
}
long long mul(long long x, long long y) { return (x * y) % mod; }
long long inv(long long x) { return power(x, mod - 2, mod); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
long long fact[Nfact];
void precalc() {
  fact[0] = 1;
  for (long long i = 1; i < Nfact; i++) fact[i] = mul(fact[i - 1], i);
}
long long nCr(long long n, long long k) {
  if (n < k) return 0;
  return divide(fact[n], mul(fact[k], fact[n - k]));
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long sum = a + b;
  cout << sum << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}
