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>
using namespace std;
typedef long long ll;
typedef pair<long long, long long> P;
struct edge{long long to,cost;};
const int inf = 1 << 27;
const long long INF = 1LL << 60;
const int COMBMAX = 1001001;
const long long MOD = 1000000007;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for(const auto& e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (long long)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (long long)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
long long mpow(long long a, long long n, long long mod = MOD){long long res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void yn(bool j){cout << (j ? "Yes" : "No") << endl; return;}
template<class Head> void pt(Head&& head){cout << head << endl; return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void debug(T v){rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> long long bcount(T1 v, T2 a){return upper_index(v, a) - lower_index(v, a);} 
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}
template<class T1, class T2> void sortTwoVecter(vector<T1> &f, vector<T2> &s, bool isReverse = false){
    vector<pair<T1, T2>> p;
    rep(i, f.size()) p.emplace_back(make_pair(f[i], s[i]));
    if (!isReverse) sort(p.begin(), p.end());
    else sort(p.rbegin(), p.rend());
    rep(i, f.size()) tie(f[i], s[i]) = p[i];
    return;
}

struct mint {
long long x;
    mint(long long x = 0):x((x % MOD + MOD) % MOD){}
    mint& operator += (const mint a) {if ((x += a.x) >= MOD) x -= MOD; return *this;}
    mint& operator -= (const mint a) {if ((x += MOD-a.x) >= MOD) x -= MOD; return *this;}
    mint& operator *= (const mint a) {(x *= a.x) %= MOD; return *this;}
    mint operator + (const mint a) const {mint res(*this); return res += a;}
    mint operator - (const mint a) const {mint res(*this); return res -= a;}
    mint operator * (const mint a) const {mint res(*this); return res *= a;}
    mint pow(long long t) const {if (!t) return 1; mint a = pow(t >> 1); a *= a; if (t&1) a *= *this; return a;}
    mint inv() const {return pow(MOD - 2);}
    mint& operator /= (const mint a) {return (*this) *= a.inv();}
    mint operator / (const mint a) const {mint res(*this); return res /= a;}
};

class combination {
    public:
    vector<mint> fact, ifact;
    combination(int n):fact(n + 1), ifact(n + 1) {
        assert(n < MOD);
        fact[0] = 1;
        for(int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for(int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        if (COMBMAX < n){
            mint ret = 1;
            for(long long i = 1; i <= k; i++){
                ret *= (n - i + 1);
                ret /= i;
            }
            return ret;
        }
        return fact[n] * ifact[k] * ifact[n - k];
    }

    mint multi(long long n, long long sum, long long l, long long r){
        long long m = r - l + 1;
        long long t = sum - n * (l - 1);
        mint ans = 0;
        for(long long k = 0; k <= (t - n) / m; k++){
            long long sign = k % 2 == 0 ? 1 : -1;
            mint temp = this->operator() (n, k);
            temp *= this->operator() (n + (t - n - m * k - 1), t - n - m * k);
            temp *= sign;
            ans += temp;
        }
        return ans;
    }
} com(COMBMAX);


map<long long, long long> p_fact(long long n){
    map<long long, long long> h;
    long long div = 2;
    while(div * div <= n){
        if (n % div == 0){
            h[div]++;
            n /= div;
            div = 1;
        }
        div++;
    }
    if (n != 1) h[n]++;
    return h;
}

int main(){
    ll N, X; cin >> N >> X;
    vector<ll> a(N), b(N - 1); 
    rep(i, N) cin >> a[i];
    rep(i, N - 1) b[i] = a[i] + a[i + 1];
    ll ans = 0;
    rep(i, N - 1){
        ll diff = max(b[i] - X, 0LL);
        ans += diff;
        if (i < N - 2) b[i + 1] -= diff;
    }
    pt(ans);



}