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 <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <vector>
#include <utility>
#include <functional>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <deque>
#include <ctime>

using namespace std;

#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define pb push_back
#define mp make_pair
#define all(r) (r).begin(),(r).end()
#define rall(r) (r).rbegin(),(r).rend()
#define fi first
#define se second
#define ARRAY_LENGTH(array) (sizeof(array) / sizeof(array[0]))


typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9;
const ll LINF = 1e18;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;
const double PI = acos(-1);

int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};

ll a[int(1e5)+5];

int main(){

    ll n,x;
    ll cnt=0;
    cin >> n >> x;
    for(ll i=0; i<n; i++) cin >> a[i];

    for(ll j=0; j<n-1; j++){
        if(a[j] + a[j+1] <= x) continue;
        if(a[j] > a[j+1]){
            ll tmp_sum = a[j] + a[j+1];
            cnt += tmp_sum - x; //総和-xが回数
            a[j] -= (tmp_sum - x); //その分,a[j]からひく
            if(a[j] < 0) {
                a[j+1] -= abs(a[j]);
                cnt += abs(a[j]); //マイナス分
            } //0未満になった分をa[j+1]から引く

        }else if(a[j] <= a[j+1]){
            ll tmp_sum = a[j] + a[j+1];
            cnt += tmp_sum - x;
            a[j+1] -= (tmp_sum - x);
            if(a[j+1] < 0) {
                a[j+1] = 0; //マイナスになられては困るので
            } //同様
        }
    }

    cout << cnt << endl;



}