To solve this problem, find a prime number greater than \(10^9\). If the input contains 1, then there is no solution.

#include <bits>
using namespace std;

typedef long long ll;

void process() {
    int size;
    cin >> size;
    bool valid = true;
    vector<int> data(size);
    
    for (int i = 0; i < size; ++i) {
        cin >> data[i];
        if (data[i] == 1) valid = false;
    }
    
    if (!valid) {
        cout << "-1\n";
        return;
    }
    cout << "1000000007\n";
}

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

    int cases;
    cin >> cases;
    while (cases--) {
        process();
    }

    return 0;
}
</int></bits>

Problem B - Unbroken Blade

This problem requires determining if a given tree can form a Hamiltonian path. This is possible only if the tree forms a chain, so check node degrees.

#include <bits>
using namespace std;

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

    int nodes;
    cin >> nodes;
    vector<int> degree(nodes + 1, 0);

    for (int i = 1; i < nodes; ++i) {
        int x, y;
        cin >> x >> y;
        degree[x]++;
        degree[y]++;
        if (degree[x] > 2 || degree[y] > 2) {
            cout << "-1\n";
            return 0;
        }
    }

    vector<int> terminals;
    for (int i = 1; i <= nodes; ++i)
        if (degree[i] == 1) terminals.push_back(i);

    cout << terminals[0] << " " << terminals.back() << "\n";
    return 0;
}
</int></int></bits>

Problem C - Diligent Shift

The strategy involves moving all '1's to the top-left corner by first pushing them upwards and then leftwards, followed by adjusting any remaining '1's towards the center.

#include <bits>
using namespace std;

int main() {
    int test_cases = 1;
    cin >> test_cases;
    while (test_cases--) {
        // Implementation details omitted for brevity
    }
    return 0;
}
</bits>

Problem D - Twin Decision

Check if the array elements can be divided into exactly two disstinct sets of equal sizes.

#include <bits>
using namespace std;

typedef long long ll;

void evaluate() {
    int length;
    cin >> length;
    set<int> unique_vals;
    vector<int> arr(length);
    
    for (int &val : arr) {
        cin >> val;
        unique_vals.insert(val);
    }
    
    if ((length % 2) != 0 || unique_vals.size() != 2) {
        cout << "No\n";
        return;
    }
    
    int count_first = count(arr.begin(), arr.end(), *unique_vals.begin());
    int count_second = count(arr.begin(), arr.end(), *next(unique_vals.begin()));
    cout << (count_first == count_second ? "Yes\n" : "No\n");
}

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

    int tests;
    cin >> tests;
    while (tests--) evaluate();

    return 0;
}
</int></int></bits>

Problem E - Twin Mistake

Calculate the minimal operations needed to convert half the array to one value and the other half to another, considering the median for optimal transformation.

#include <bits>
using namespace std;

int main() {
    int scenarios = 1;
    cin >> scenarios;
    while (scenarios--) {
        // Simplified implementation based on original concept
    }
    return 0;
}
</bits>

Problem G - Balanced Order

Insure the sum of elemments equals the sum of indices from 1 to n. Then, sort the array and calculate the required adjustments.

#include <bits>
using namespace std;

typedef long long ll;

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

    int elements;
    cin >> elements;
    ll total_sum = 0;
    vector<int> values(elements + 1);

    for (int i = 1; i <= elements; ++i) {
        cin >> values[i];
        total_sum += values[i];
    }

    if (total_sum != (ll)elements * (elements + 1) / 2) {
        cout << "-1\n";
        return 0;
    }

    sort(values.begin() + 1, values.end());
    ll correction = 0;
    for (int i = 1; i <= elements; ++i)
        if (values[i] < i) correction += i - values[i];

    cout << correction << "\n";
    return 0;
}
</int></bits>

Problem H - Window Order

Sort intervals by their right endpoints, then assign each interval a unique number ensuring it fits within its bounds.

#include <bits>
using namespace std;

typedef long long ll;

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

    int intervals;
    cin >> intervals;
    set<int> available;
    vector<array>> ranges(intervals);

    for (int i = 0; i < intervals; ++i) {
        cin >> ranges[i][0] >> ranges[i][1];
        ranges[i][2] = i + 1;
        available.insert(i + 1);
    }

    sort(ranges.begin(), ranges.end(), [](auto& a, auto& b) {
        return a[1] == b[1] ? a[0] > b[0] : a[1] < b[1];
    });

    vector<int> results(intervals + 1);
    for (auto &[start, end, idx] : ranges) {
        auto pos = available.lower_bound(start);
        if (pos != available.end() && *pos <= end) {
            results[idx] = *pos;
            available.erase(pos);
        } else {
            cout << "-1\n";
            return 0;
        }
    }

    for (int i = 1; i <= intervals; ++i)
        cout << results[i] << (i < intervals ? " " : "\n");

    return 0;
}
</int></array></int></bits>