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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Practice
{
    class Program
    {
        static void Main(string[] args)
        {
            Solve();
        }
        private const long MOD = 1000000007;
        
        static void Solve()
        {
            var o = ReadAndParseIntArr();
            int n = o[0];
            int x = o[1];
            var a = ReadAndParseLongArr();

            long ans = GetProcTotal(a, x);
            Console.WriteLine(ans);
        }
        private static long GetProcTotal(long[] b, long x)
        {
            var a = new long[b.Length];
            Array.Copy(b, a, b.Length);
            long ans = 0;
            for (long i = 0; i < a.Length - 1; ++i)
            {
                long t = a[i] + a[i + 1];
                if (t > x)
                {
                    long sub = t - x;
                    ans += sub;

                    if (a[i] > x)
                    {
                        
                        a[i] = x - a[i+1];
                        a[i + 1] = 0;
                    }
                    else
                    {
                        a[i + 1] -= sub;
                    }
                }
            }
            return ans;
        }

        #region よく使う便利関数
        private static bool isPrime(long x)
        {
            if (x == 2) { return true; }
            if (x < 2 || x % 2 == 0) { return false; }
            long i = 3;
            while (i * i <= x)
            {
                if (x % i == 0) { return false; }
                i = i + 2;
            }
            return true;
        }
        private static long lcm(long x, long y)
        {
            return x / gcd(x, y) * y;
        }
        private static long gcd(long x, long y)
        {
            return y > 0 ? gcd(y, x % y) : x;
        }
        private static long pow(long x, long n)
        {
            if (n == 0) { return 1; }
            long res = pow(x * x % MOD, n / 2);
            if (n % 2 == 1)
            {
                res = res * x % MOD;
            }
            return res;
        }
        private static int ReadAndParseInt()
        {
            return int.Parse(Console.ReadLine());
        }
        private static int[] ReadAndParseIntArr()
        {
            return Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
        }
        private static long ReadAndParseLong()
        {
            return long.Parse(Console.ReadLine());
        }
        private static long[] ReadAndParseLongArr()
        {
            return Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
        }
        #endregion
    }
}