human
JAVA solution for "Recently, Petya learned about a new game "Slay the Dragon". As the name suggests, the player will have to fight with dragons. To defeat a dragon, you have to kill it and defend your castle. To do this, the player has a squad of n heroes, the strength of the i-th hero is equal to a_i.

According to the rules of the game, exactly one hero should go kill the dragon, all the others will defend the castle. If the dragon's defense is equal to x, then you have to send a hero with a strength of at least x to kill it. If the dragon's attack power is y, then the total strength of the heroes defending the castle should be at least y.

The player can increase the strength of any hero by 1 for one gold coin. This operation can be done any number of times.

There are m dragons in the game, the i-th of them has defense equal to x_i and attack power equal to y_i. Petya was wondering what is the minimum number of coins he needs to spend to defeat the i-th dragon.

Note that the task is solved independently for each dragon (improvements are not saved).

Input

The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — number of heroes.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^{12}), where a_i is the strength of the i-th hero.

The third line contains a single integer m (1 ≤ m ≤ 2 ⋅ 10^5) — the number of dragons.

The next m lines contain two integers each, x_i and y_i (1 ≤ x_i ≤ 10^{12}.
1 ≤ y_i ≤ 10^{18}) — defense and attack power of the i-th dragon.

Output

Print m lines, i-th of which contains a single integer — the minimum number of coins that should be spent to defeat the i-th dragon.

Example

Input

4
3 6 2 3
5
3 12
7 9
4 14
1 10
8 7

Output

1
2
4
0
2

Note

To defeat the first dragon, you can increase the strength of the third hero by 1, then the strength of the heroes will be equal to [3, 6, 3, 3]. To kill the dragon, you can choose the first hero.

To defeat the second dragon, you can increase the forces of the second and third heroes by 1, then the strength of the heroes will...":



Give me a Python code to print m lines, i-th of which contains a single integer — the minimum number of coins that should be spent to defeat the i-th dragon?
assistant
#Here is Python code to print m lines, i-th of which contains a single integer — the minimum number of coins that should be spent to defeat the i-th dragon.
 import java.util.*;
import java.io.*;

public class C {
    static MyScanner sc;
    static PrintWriter out;
    static {
        sc = new MyScanner();
        out = new PrintWriter(System.out);
    }
    public static void solve(TreeSet<Long> set, long sum, long x, long y) {
        long def;
        long ans = Long.MAX_VALUE;
        Object temp;
        temp = set.floor(x);
        if(temp != null) {
            long h = (long)temp;
            long t = 0l;
            if(h < x) t += x - h;
            def = sum - h;
            if(def < y) t += y - def;
            ans = Math.min(ans, t);
            if(h == x) {
                out.println(ans);
                return;
            }
        }
        temp = set.ceiling(x);
        if(temp != null) {
            long h = (long)temp;
            long t = 0l;
            if(h < x) t += x - h;
            def = sum - h;
            if(def < y) t += y - def;
            ans = Math.min(ans, t);
        }
        out.println(ans);
    }
    public static void main(String[] args) {
        int n = sc.nextInt();
        long s = 0;
        TreeSet<Long> set = new TreeSet<>();
        for(int i = 0; i < n; i++) {
            long x = sc.nextLong();
            set.add(x);
            s += x;
        }
        int t = sc.nextInt();
        while(t-- > 0) {
            long x = sc.nextLong();
            long y = sc.nextLong();
            solve(set, s, x, y);
        }
        out.flush();
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer tok;
    MyScanner() {
        try { br = new BufferedReader(new InputStreamReader(System.in)); }
        catch(Exception e) { System.out.println(e); }
        tok = new StringTokenizer("");
    }
    public String next() {
        try {
            while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
        }
        catch(Exception e) { System.out.println(e); }
        return tok.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long...