function getMoneySpent(keyboards, drives, b) {
    if(Math.min(...keyboards)+Math.min(...drives) > b){
        return -1
    }
    
    const inBudget = [];
    
    for(let i = 0; i < keyboards.length; i++){
        for(let j = 0; j < drives.length; j++){
            const sum = keyboards[i]+drives[j];
            
            if(sum <= b){
                inBudget.push(sum);
            }
        }
    }
    
    return Math.max(...inBudget);
}

def getMoneySpent(keyboards, drives, b):
    l1=[]
    for i in keyboards:
        for j in drives:
            if i+j<=b:
                l1.append(i+j)
    if len(l1)==0:return -1
    return max(l1)

//my js solution

function getMoneySpent(keyboards, drives, b) {
    let sum = 0;
    let budget = []
    for (let i = 0; i < keyboards.length; i++) {
        for (let j = 0; j < drives.length; j++) {
            sum = keyboards[i] + drives[j]
            if (sum <= b) {
                budget.push(sum)
            }
        }
    }
    let result = Math.max(...budget)
    if (budget.length === 0) {
        return -1
    } else {
        return result
    }

}

function getMoneySpent(keyboards, drives, b) {
    /*
     * I know the solution is pretty straight forward, but I like my formating the best
     */
    let mostExpensivePurchase = -1;
    for (let i = 0; i < keyboards.length; i++){
        for (let j = 0; j < drives.length; j++){
            let sum = keyboards[i] + drives[j];
            if(sum <= b && sum > mostExpensivePurchase) mostExpensivePurchase = sum;
        }
    }
    return mostExpensivePurchase;

}

function getMoneySpent(keyboards, drives, b) {
    let m = keyboards.length;
    let n = drives.length;
    let val = 0;
    for(let i = 0; i < m; i++){
        for(let j = 0; j < n; j++){
            if(keyboards[i]+drives[j] <= b && keyboards[i]+drives[j] > val){
                val = keyboards[i]+drives[j];
            }
        }
    }
    if(val == 0){
        return -1;
    }
    else{