text: such car that Vasya will be able to reach the cinema before the film starts (not later than in t minutes).If there is no appropriate car, print -1.Examples Input 3 1 8 10 10 8 5 7 11 9 3 Output 10 Input 2 2 10 18 10 4 20 6 5 3 Output 20 Note In the first sample, Vasya can reach the cinema in time using the first or the third cars, but it would be cheaper to choose the first one.Its price is equal to 10, and the capacity of its fuel tank is 8.Then Vasya can drive to the first gas station in the accelerated mode in 3 minutes, spending 6 liters of fuel.After that he can full the tank and cover 2 kilometers in the normal mode in 4 minutes, spending 2 liters of fuel.Finally, he drives in the accelerated mode covering the remaining 3 kilometers in 3 minutes and spending 6 liters of fuel.# Solution import java.awt.*; import java.io.*; import java.math.BigDecimal; import java.math.BigInteger; import java.util.*; import java.util.List; import static java.lang.Math.max; import static java.lang.Math.min; public class A implements Runnable{ // SOLUTION!!!// HACK ME PLEASE IF YOU CAN!!!// PLEASE!!!// PLEASE!!!// PLEASE!!!private final static Random rnd = new Random(); private final static String fileName = ""; private static class Car { int cost; int volume; public Car(int cost, int volume) { this.cost = cost; this.volume = volume; } } private void solve() { int carsCount = readInt(); int fuelsCount = readInt(); int distance = readInt(); int time = readInt(); Car[] cars = new Car[carsCount]; for (int i = 0; i < carsCount; ++i) { cars[i] = new Car(readInt(), readInt()); } int[] fuels = new int[fuelsCount + 1]; for (int i = 0; i < fuelsCount; ++i) { fuels[i] = readInt(); } fuels[fuelsCount] = distance; Arrays.sort(fuels); int answer = getAnswer(cars, fuels, distance, time); out.