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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define INF 2000000000
#define int long long int
#define MOD 1000000007
#define pb push_back
#define cin1(n) (cin >> n)
#define cin2(a,b) (cin >> a >> b)
#define cin3(a,b,c) (cin >> a >> b >> c)
#define cout(n) (cout << n << endl)
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define fr first 
#define sc second 
#define EPS (1e-10)

using namespace std;

typedef vector<int> vi;
typedef vector< vector<int> > vvi;
typedef vector<pair<int,int>> vpii;
typedef vector<string> vs;
typedef vector<vector<string>> vss;
typedef pair<int,int> pii;

int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
int n,m;
int a,b;
int p,q;


signed main() {
	cin1(n);
	int x;cin1(x);

	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin1(a[i]);
	}

	int ans = 0;
	for (int i = 0; i < n-1; i++) {
		int sum = a[i] + a[i+1];
		if (sum > x) {
			int eat = sum - x;
			ans += eat;
			a[i+1] -= eat;
			/*
			if (a[i+1] < 0) {
				a[i] += a[i+1];
				a[i+1] = 0;
			}
			*/
		}
		//cout << a[i] << endl;
	}
	cout(ans);

	return 0;
}