SELECT ID, NAME
FROM CUSTOMER
WHERE COUNTRY = 'USA' AND CREDITS > 100000
ORDER BY ID;

#include <bits/stdc++.h>
using namespace std;

void solve(int N, int M, vector<pair<int, int>>& a, int q, vector<int>& queries) {
    vector<vector<int>> adj(N + 1);

    for (int i = 0; i < M; ++i) {
        int u = a[i].first;
        int v = a[i].second;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    for (int i = 1; i <= N; ++i) {
        sort(adj[i].begin(), adj[i].end());
    }

    for (int i = 0; i < q; ++i) {
        int prof = queries[i];

        if (adj[prof].empty()) {
            cout << -1 << endl;
        } else {
            for (int nbr : adj[prof]) {
                cout << nbr << " ";
            }
            cout << endl;
        }
    }
}

from collections import defaultdict
def countOfAtoms(formula: str) -> str:
    def parse_formula(i):
        n = len(formula)
        count = defaultdict(int)
        while i < n:
            if formula[i] == '(':
                i, inner_count = parse_formula(i + 1)
                multiplier = 0
                i += 1
                while i < n and formula[i].isdigit():
                    multiplier = multiplier * 10 + int(formula[i])
                    i += 1
                multiplier = multiplier or 1
                for elem in inner_count:
                    count[elem] += inner_count[elem] * multiplier
            elif formula[i] == ')':
                return i, count
            else:
                j = i + 1
                while j < n and formula[j].islower():
                    j += 1
                element = formula[i:j]
                i = j
                multiplier = 0
                while j < n and formula[j].isdigit():
                    multiplier = multiplier * 10 + int(formula[j])
                    j += 1
                count[element] += max(multiplier, 1)
                i = j
        return i, count

    _, counts = parse_formula(0)
    sorted_elements = sorted(counts.items())
    result = []
    for element, count in sorted_elements:
        result.append(element)
        if count > 1:
            result.append(str(count))
    return ''.join(result)