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: Runtime Error)

//#define _GRIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

// 基本テンプレート
#pragma region MACRO
#define P(x) cout << (x) << endl
#define p(x) cout << (x)
#define PED cout << "\n"
#define rep(i,n) for(int i=0; i<(int)n; ++i)
#define REP(i,x,n) for(int i=x; i<(int)n; ++i)
#define repi(i,n) for(int i=0; i<=(int)n; ++i)
#define REPI(i,x,n) for(int i=x; i<=(int)n; ++i)
#define ILP while(true)
#define FOR(i,c) for(__typeof((c).begin())!=(c).begin(); i!=(c).end(); ++i)
#define ALL(c) (c).begin(), (c).end()
#define mp make_pair
#pragma endregion

// 型
#pragma region TYPE_DEF
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<string, string> pss;
typedef pair<string, int> psi;
typedef pair<int, string> pis;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long double> vld;
typedef vector<long> vl;
typedef vector<long long> vll;
typedef vector<string> vs;
#pragma endregion

// Effective std
#pragma region ESTD
template<typename C, typename T> int count(C& c, T t) { return count(ALL(c), t); }
template<typename C, typename F> int count_if(C& c, F f) { return count_if(ALL(c), f); }
template<typename C, typename T> void erase(C& c, T t) { remove(ALL(c), t), c.end(); }
template<typename C> void remove(vector<C>& c, unsigned int index) { c.erase(c.begin()+index); }
template<typename C, typename T, typename U> void replace(C& c, T t, U u) { replace(ALL(c), t, u); }
template<typename C, typename F, typename U> void replace_if(C& c, F f, U u) { (ALL(c), f, u); }
template<typename C> void reverse(C& c) { reverse(ALL(c)); }
template<typename C> void sort(C& c) { sort(ALL(c)); }
template<typename C, typename Pred> void sort(C& c, Pred p) { sort(ALL(c), p); }
#pragma endregion

// 定数
#pragma region CONST_VAL
constexpr int PI = (2*acos(0.0));
constexpr int EPS = (1e-9);
constexpr int MOD = (int)(1e9+7);
constexpr int INF = 100000000;
#pragma endregion

int N, x;
vi a;

int main()
{
    cin >> N >> x;
    rep(i,N) {
        int m;
        cin >> m;
        a.push_back(m);
    }
    vi a2 = a;
    int ans1 = 0;
    rep(i,N-1) {
        if(a[i]+a[i+1]>x) {
            int t = a[i]+a[i+1]-x;
            ans1 += t;
            rep(i,t) {
                if(a[i+1]>0) a[i+1] -= 1;
                else a[i] -= 1;
            }
        }
    }
    int ans2 = 0;
    rep(i,N-1) {
        if(a2[N-1-i]+a2[N-2-i]>x) {
            int t = a2[N-1-i]+a2[N-2-i]-x;
            ans2 += t;
            rep(i,t) {
                if(a2[N-2-i]>0) a2[N-2-i] -= 1;
                else a[N-1-i] -= 1;
            }
        }
    }
    P(min(ans1, ans2));
    return 0;
}