[TASK]:
You will be given lines of pseudocode, your task is to write the corresponding C++ code. The pseudocode will provide detailed description of the c++ code line by line. The pseudocode is garanteed to be correct and complete.

[INSTRUCTION]:
The following libraries are already included in the code.
```cpp
#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <fstream>
#include <climits>
#include <cassert>
#include <iomanip>
#include <sstream>
#include <bitset>
using namespace std;
```
Do not include them in your code again. Please surround your code with ```cpp and ``` markers. Note that the code should correspond to the pseudocode line by line.

[PSEUDOCODE]:
create int array d of size 30
create integer set ma
define the main function
    make ints a, b, l, and r
    read a, b, l, and r
    if a is equal to 3 and b is equal to 1 and l is equal to 4 and r is equal to 10
        print 4
        return to indicate successful execution
    end of the block
    make string s
    for i = 1 to a, add i - 1 + 'a' to s
    let int x = s.size() - 1
    for i = 1 to b, set s to s + s[x]
    set x to s.size() - 1
    for i = 1 to a
        set d[s[x] - 'a' to 1
        decrease x by 1
    end of the block
    for i = 1 to a
        for j = 0 to 26 exclusive
            if d[j] is equal to 0
                set s to s + j + 'a'
                set d[j] to 1
                break loop
            end of the block
        end of the block
    end of the block
    set x to s.size() - 1
    for i = 1 to b, add s[x] to s
    if r is less than or equal to 2 * (a + b)
        for i = l - 1 to r - 1, insert s[i] - 'a' to max
    else do
        create int xx = l / (2 * a + 2 * b)
        let int yy = r / (2 * a + 2 * b)
        if xx is yy
            set l to l % (2 * a + 2 * b)
            set r to r % (2 * a + 2 * b)
            for i = l - 1 to r exclusive, insert s[i] - 'a' into ma
        else if yy is equal to xx + 1
            set l to l % (2 * a + 2 * b)
            set r to r % (2 * a + 2 * b)
            for i = l - 1 to size of s exclusive, insert s[i] - 'a' into ma
            for i = 0 to r exclusive, insert s[i] - 'a' into ma
        else do
            for i = 0 to s.size() exclusive, insert s[i] - 'a' into ma
        end of the block
    end of the block
    print ma.size()
end of the block

[CODE]:

int d[30];
set<int> ma;
int main() {
    int a, b, l, r;
    cin >> a >> b >> l >> r;
    if (a == 3 && b == 1 && l == 4 && r == 10) {
        cout << 4 << endl;
        return 0;
    }
    string s;
    for (int i = 1; i <= a; i++) { s += i - 1 + 'a'; }
    int x = s.size() - 1;
    for (int i = 1; i <= b; i++) s += s[x];
    x = s.size() - 1;
    for (int i = 1; i <= a; i++) {
        d[s[x] - 'a'] = 1;
        x--;
    }
    for (int i = 1; i <= a; i++) {
        for (int j = 0; j < 26; j++) {
            if (d[j] == 0) {
                s += j + 'a';
                d[j] = 1;
                break;
            }
        }
    }
    x = s.size() - 1;
    for (int i = 1; i <= b; i++) s += s[x];
    if (r <= 2 * (a + b)) {
        for (int i = l - 1; i <= r - 1; i++) { ma.insert(s[i] - 'a'); }
    } else {
        int xx = l / (2 * a + 2 * b);
        int yy = r / (2 * a + 2 * b);
        if (xx == yy) {
            l = l % (2 * a + 2 * b);
            r = r % (2 * a + 2 * b);
            for (int i = l - 1; i < r; i++) ma.insert(s[i] - 'a');
        } else if (yy == xx + 1) {
            l = l % (2 * a + 2 * b);
            r = r % (2 * a + 2 * b);
            for (int i = l - 1; i < s.size(); i++) ma.insert(s[i] - 'a');
            for (int i = 0; i < r; i++) ma.insert(s[i] - 'a');
        } else {
            for (int i = 0; i < s.size(); i++) ma.insert(s[i] - 'a');
        }
    }
    cout << ma.size() << endl;
}