Problem:
Score : 300 points

Problem Statement

There are N boxes arranged in a row.
Initially, the i-th box from the left contains a_i candies.

Snuke can perform the following operation any number of times:

Choose a box containing at least one candy, and eat one of the candies in the chosen box.

His objective is as follows:

Any two neighboring boxes contain at most x candies in total.

Find the minimum number of operations required to achieve the objective.

Constraints

2 ≤ N ≤ 10^5

0 ≤ a_i ≤ 10^9

0 ≤ x ≤ 10^9

Input

The input is given from Standard Input in the following format:

N x
a_1 a_2 ... a_N

Output

Print the minimum number of operations required to achieve the objective.

Sample Input 1

3 3
2 2 2

Sample Output 1

1

Eat one candy in the second box.
Then, the number of candies in each box becomes (2, 1, 2).

Sample Input 2

6 1
1 6 1 2 0 4

Sample Output 2

11

For example, eat six candies in the second box, two in the fourth box, and three in the sixth box.
Then, the number of candies in each box becomes (1, 0, 1, 0, 0, 1).

Sample Input 3

5 9
3 1 4 1 5

Sample Output 3

0

The objective is already achieved without performing operations.

Sample Input 4

2 0
5 5

Sample Output 4

10

All the candies need to be eaten.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	void run() {
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		int x = sc.nextInt();

		int[] a=new int[N];
		int[] init=new int[N];

		//初期の値を記録しておく。
		for (int i = 0; i < N; i++) {
			a[i]=sc.nextInt();
			init[i]=a[i];
		}

		//左端をx以下を保証。
		a[0]=Math.min(a[0],x);

		//a[i]+a[i-1]がx以下になるには
		//a[i]をx-a[i-1]まで減らしてあげるといい

		for (int i = 1; i < N; i++) {
			a[i]=Math.min(a[i],x-a[i-1]);
		}

		long ans=0;
		//初期の値との差を足し上げていく。
		for (int i = 0; i < N; i++) {
			ans+=init[i]-a[i];
		}

		System.out.println(ans);
	}

	//以下テンプレート

	public static int[] extgcd(int a, int b) {

		int x0 = 1;
		int x1 = 0;

		int y0 = 0;
		int y1 = 1;

		while (b != 0) {
			int q = a / b;
			int r = a % b;

			int x2 = x0 - q * x1;
			int y2 = y0 - q * y1;

			a = b;
			b = r;

			x0 = x1;
			x1 = x2;

			y0 = y1;
			y1 = y2;
		}

		return new int[] { a, x0, y0 };

	}

	static int gcd(int a, int b) {

		if (b == 0)
			return a;

		if (a < b) {
			int t = a;
			a = b;
			b = t;
		}

		return gcd(b, a % b);

	}

	static int lcm(int a, int b) {
		return a * b / gcd(a, b);
	}

	static void swap(int[] a) {
		int t;

		t = a[0];
		a[0] = a[1];
		a[1] = t;

		return;
	}

	static <T> void output(List<T> list) {

		for (int i = 0; i < list.size(); i++) {
			System.out.print(list.get(i));

			if (i != list.size() - 1) {
				System.out.print(" ");
			} else {
				nl();
			}
		}

	}

	static void output(String[][] str) {

		for (int i = 0; i < str.length; i++) {
			for (int j = 0; j < str[i].length; j++) {
				print(str[i][j]);
			}

			nl();
		}

	}

	static void output(boolean flg, String Yes, String No) {

		if (flg) {
			pln(Yes);
		} else {
			pln(No);
		}

	}

	static void output(String[][] str, int digit) {

		String dig = "%" + String.valueOf(digit) + "s";

		for (int i = 0; i < str.length; i++) {
			for (int j = 0; j < str[i].length; j++) {
				System.out.printf(dig, str[i][j]);
			}
			nl();
		}

	}

	static void pln(String str) {
		System.out.println(str);
	}

	static void pln(int x) {
		System.out.println(x);
	}

	static void print(String str) {
		System.out.print(str);
	}

	static void print(int x) {
		System.out.print(x);
	}

	static void print(String str, int times) {
		for (int i = 0; i < times; i++) {
			print(str);
		}
	}

	static void print(int x, int times) {
		for (int i = 0; i < times; i++) {
			print(x);
		}
	}

	static void nl() {
		System.out.println();
	}

}