The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.

If x is the number of passengers in a bus just before the current bus stop and y is the number of passengers in the bus just after current bus stop, the system records the number y-x. So the system records show how number of passengers changed.

The test run was made for single bus and n bus stops. Thus, the system recorded the sequence of integers a_1, a_2, ..., a_n (exactly one number for each bus stop), where a_i is the record for the bus stop i. The bus stops are numbered from 1 to n in chronological order.

Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w (that is, at any time in the bus there should be from 0 to w passengers inclusive).

Input

The first line contains two integers n and w (1 ≤ n ≤ 1 000, 1 ≤ w ≤ 10^{9}) — the number of bus stops and the capacity of the bus.

The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} ≤ a_i ≤ 10^{6}), where a_i equals to the number, which has been recorded by the video system after the i-th bus stop.

Output

Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.

Examples

Input

3 5
2 1 -3


Output

3


Input

2 4
-1 1


Output

4


Input

4 10
2 4 1 2


Output

2

Note

In the first example initially in the bus could be 0, 1 or 2 passengers.

In the second example initially in the bus could be 1, 2, 3 or 4 passengers.

In the third example initially in the bus could be 0 or 1 passenger.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EBusVideoSystem solver = new EBusVideoSystem();
        solver.solve(1, in, out);
        out.close();
    }

    static class EBusVideoSystem {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            int n = in.nextInt();
            int w = in.nextInt();

            in.nextLine();

            long[] a = new long[n];

            for (int i = 0; i < n; i++) {
                a[i] = in.nextLong();
            }

            Long relMin = 0l;
            Long relMax = 0l;
            Long relPassangers = 0L;
            for (int i = 0; i < n; i++) {
//            System.out.println("relMin: " + relMin +", relMax: " + relMax);
                relPassangers += a[i];
                relMin = Math.min(relMin, relPassangers);
                relMax = Math.max(relMax, relPassangers);
            }

//        System.out.println("final relMin: " +relMin + ", final relMax: " + relMax);

            //check if relMin and relMax "diff" is between 0 and w
            boolean posMinPosMax = relMax >= 0l && relMin >= 0l;
            boolean negMinPosMax = relMin < 0l && relMax >= 0l;
            boolean negMinNegMax = relMin < 0l && relMax < 0l;

            long diff;
            if (posMinPosMax)
                diff = relMax - relMin;
            else if (negMinPosMax)
                diff = relMax - relMin;
            else
                diff = Math.abs(relMin) - Math.abs(relMax);


            if (diff < 0 || diff > w) {
                out.println(0);
                return;
            }

            //check possibilities
            int MAX = w;
            int MIN = 0;

            int ans = 1;

            ans += (MAX - relMax);
            ans += (relMin - MIN);

            out.println(ans);
        }

    }
}

