<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

100. A+B

time limit per test: 0.25 sec.
memory limit per test: 65536 KB


Read integers A and B from input file and write their sum in output file.


Input

Input file contains A and B (0<A,B<10001).


Output

Write answer in output file.


Sample Input

5 3
Sample Output

8

<|response|>
1. Abridged problem statement
Given two positive integers A and B (1 ≤ A, B ≤ 10000), compute and output A + B.

2. Key observations needed to solve the problem
- A and B are at most 10000, so A + B ≤ 20000, which easily fits in a standard 32-bit signed integer.
- There are only two numbers to process, so any reasonable I/O method will run in constant time and use negligible memory.
- No special edge cases (like zero or negative inputs) occur, since 0 < A, B < 10001.

3. Full solution approach based on the observations
a. Read two integers from the input.
b. Compute their sum with the built-in integer addition operator.
c. Print the result.
d. Since the input size is fixed and tiny, the time and memory complexity are both O(1).

4. C++ implementation
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int a, b;

void read() { cin >> a >> b; }

void solve() {
    // Read two integers and print their sum.

    cout << a + b << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python implementation with detailed comments
```python
# 1. Read a line from input, split it into two tokens, and convert each to int
A, B = map(int, input().split())

# 2. Compute the sum of the two integers
result = A + B

# 3. Print the result
print(result)
```
