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: Accepted)

#include <algorithm>
#include <cmath>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using ll = long long;
using namespace std;
using P = pair<ll, ll>;
static ll imin = std::numeric_limits<ll>::min() / 10;
static ll imax = std::numeric_limits<ll>::max() / 10;
#define rep(i, n) for (ll i = 0; (i) < (n); ++(i))
#define rep1(i, n) for (ll i = 1; (i) < (n); ++(i))
#define rev_rep(i, n) for (ll i = (n)-1; (i) >= 0; --(i))
#define range(a, i, b) for (ll i = (a); (i) < (b); ++(i))

#ifdef LOCAL
#define dump(x) cout << #x << ":" << x << "\n";
#endif
#ifndef LOCAL
#define dump(x)
#endif

#define all(v) v.begin(), v.end()
#define readl(N) \
  ll N;          \
  cin >> N;
#define readlv(N, vec) \
  vector<ll> vec(N);   \
  rep(i, N) { cin >> vec[i]; }
#define greater_than [](ll l, ll r) { return l > r; }
template <class T1, class T2>
std::ostream& operator<<(std::ostream& os, std::map<T1, T2>& mp);
template <class T1, class T2>
std::ostream& operator<<(std::ostream& os, std::pair<T1, T2>& pair);
template <class T>
std::ostream& operator<<(std::ostream& os, std::vector<T>& vec);
template <class T1, class T2>
std::ostream& operator<<(std::ostream& os, std::pair<T1, T2>& pair) {
  os << '(' << pair.first << ',' << pair.second << ')';
  return os;
}

template <class T>
std::ostream& operator<<(std::ostream& os, std::vector<T>& vec) {
  std::size_t n = vec.size();
  if (n == 0) {
    return os;
  }
  os << " ";
  for (ll i = 0; i < n - 1; ++i) {
    os << vec[i] << " ";
  }
  os << vec[n - 1];
  os << std::endl;
  return os;
}

template <class T1, class T2>
std::ostream& operator<<(std::ostream& os, std::map<T1, T2>& mp) {
  std::size_t n = mp.size();
  if (n == 0) {
    return os;
  }
  os << " ";
  for (auto&& x : mp) {
    os << x << " ";
  }
  os << std::endl;
  return os;
}

template <class T>
std::ostream& operator<<(std::ostream& os, std::set<T>& mp) {
  std::size_t n = mp.size();
  if (n == 0) {
    return os;
  }
  os << " ";
  for (auto&& x : mp) {
    os << x << " ";
  }
  os << std::endl;
  return os;
}

template <class T>
inline bool chmin(T& a, T const b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T const b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

ll fact(ll n) {
  ll res = 1;
  while (n > 0) {
    res *= n;
    n--;
  }
  return res;
}

ll pow(ll i, ll n) {
  ll res = 1;
  while (n >= 0) {
    res *= i;
    n--;
  }
  return res;
}

ll gcd(ll l, ll r) {
  if (l < r) {
    return gcd(r, l);
  }
  if (r == 0) {
    return l;
  }
  return gcd(r, l % r);
}

template <class T>
vector<T> accum_sum(vector<T> const& a) {
  vector<T> b(a.size() + 1);
  b[0] = 0;
  rep(i, a.size()) { b[i + 1] = b[i] + a[i]; }
  return b;
}
constexpr ll mod = 1e9 + 7;

bool isprime(ll n) {
  range(2, i, sqrt(n) + 1) {
    if (n == i) {
      break;
    }
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int main(int argc, char const* argv[]) {
  ll N, x;
  cin >> N >> x;
  vector<ll> a(N);
  rep(i, N) { cin >> a[i]; }
  dump(a);
  ll prev = 0;
  ll res = 0;
  rep(i, N) {
    if (a[i] + prev > x) {
      res += a[i] - (x - prev);
      a[i] = x - prev;
    }
    prev = a[i];
  }
  dump(a);
  cout << res << endl;
  return 0;
}