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 <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <fstream> 

using namespace std;

inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

template<class T> inline T sqr(T x) { return x*x; }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long int> vll;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define P(p) cout<<(p)<<endl;
#define VEC_2D(a,b) vector<vector<int> >(a, vector<int>(b, 0))
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define INF (1100000000)
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define MOD 1000000007LL
#define FSP(a) cout << fixed << setprecision(a)


ll gcd(ll x, ll y) {
	if (y == 0) return x;
	else return gcd(y, x%y);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}
bool is_prime(int n) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) return false;
	}
	return n != 1;
}
map<int, int> prime_factor(int n) {
	map<int, int> res;
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}
int extgcd(int a, int b, int& x, int& y) {//
	int d = a;
	if (b != 0) {
		d = extgcd(b, a%b, y, x);
		y -= (a / b)*x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}
ll mod_pow(ll x, ll n, ll mod) {
	if (n == 0) return 1;
	ll res = mod_pow(x * x % mod, n / 2, mod);
	if (n & 1) res = res * x % mod;
	return res;
}
vector<string> split(const string &str, char delim) {
	vector<string> res;
	size_t current = 0, found;
	while ((found = str.find_first_of(delim, current)) != string::npos) {
		res.push_back(string(str, current, found - current));
		current = found + 1;
	}
	res.push_back(string(str, current, str.size() - current));
	return res;
}

bool is_kadomatsu(int a, int b, int c) {
	if (a == b || a == c || b == c)return false;
	if (a > b && c > b) return true;
	if (a < b && c < b)return true;
	return false;
}


struct UF {
	int n;
	vi d;
	UF() {}
	UF(int n) :n(n), d(n, -1) {}
	int root(int v) {
		if (d[v] < 0) return v;
		return d[v] = root(d[v]);
	}
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (size(x) < size(y)) swap(x, y);
		d[x] += d[y];
		d[y] = x;
		return true;
	}
	int size(int v) { return -d[root(v)]; }
};



vector<int> divisor(int n) {
	if (n == 1) return{};
	vi res;
	for (int i = 1; i*i <= n; i++) {
		if (n%i == 0) {
			res.emplace_back(i);
			if (i != 1 && i != n / i)res.emplace_back(n / i);
		}
	}
	return res;
}
//-------------------------------------------------------------


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int a, b;
	cin >> a >> b;
	vll c(a);
	rep(i, a)cin >> c[i];
	ll ans = 0;
	for (int i = 1; i < a; i++) {
		if (c[i] + c[i - 1] > b) {
			int cie = min(c[i] + c[i - 1] - b, c[i]);

			int ci1 = c[i] + c[i - 1] - b - cie;
			c[i] -= cie;
			c[i - 1] -= ci1;
			ans += cie + ci1;
		}
	}
	P(ans);
	return 0;
		//ifstream ifs("soc-LiveJournal1.txt"); //グラフの読み込み
		//ofstream output("soc-LiveJournal2.txt");
		//int u, v;
		//string s;
		//bool is_0_start = false;
		//int count = 0;
		//rep(i, 4) getline(ifs, s);
		//for (; ifs >> u >> v;) {
		//	if (u == 0) is_0_start = true;
		//	if (!is_0_start) {
		//		u--;
		//		v--;
		//	}
		//	count++;
		//	if(count%10000==0)cout << count << endl;
		//	output << u << " " << v << endl;
		//}
}