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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <numeric>
#include <vector>
#include <string>
#include <bitset>
#include <queue>
#include <tuple>
#include <regex>
#include <cmath>
#include <map>
#include <cstring>

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

using namespace std;

using ll = long long;
using ull = unsigned long long;

using Pint = pair<int, int>;
using Pll  = pair<ll, ll>;
using Pull = pair<ull, ull>;

#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define F first
#define S second

#define rep( i, n ) for ( int i = 0; i < (int)( n ); ++i )
#define reps( i, n ) for ( int i = 1; i <= (int)( n ); ++i )
#define rrep( i, n ) for ( int i = (int)( ( n ) - 1 ); i >= 0; --i )
#define rreps( i, n ) for ( int i = (int)( ( n ) ); i > 0; --i )
#define arep( i, v ) for ( auto &&i : ( v ) )

template<typename T> T gcd( const T a, const T b ) { return ( b ? gcd( b, a % b ) : a ); }
template<typename T> T lcm( const T a, const T b ) { return ( a / gcd( a, b ) * b ); }

#define ALL( c ) ( c ).begin(), ( c ).end()
#define RALL( c ) ( c ).rbegin(), ( c ).rend()
#define UNIQUE( c ) ( c ).erase( unique( ( c ).begin(), ( c ).end() ), ( c ).end() )

constexpr ll MOD = 1000000007LL;
template<typename T = ll> constexpr T MAX = numeric_limits<T>::max();

#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev

template<typename T = ll> class UnionFind {
public:
	UnionFind( T n ) { par.resize( n ); siz.resize( n ); rep( i, n ) { par[i] = i; siz[i] = 1; } }
	T find( T x ) { if ( x == par[x] ) return ( x ); else return( par[x] = find( par[x] ) ); }
	void unite( T x, T y ) { T xx = find( x ); T yy = find( y ); if ( xx == yy ) return;
		if ( siz[xx] <= siz[yy] ) swap( xx, yy ); par[yy] = xx; siz[xx] += siz[yy]; }
private:
	vector<T> par, siz;
};

template<typename T = ll> class CompareMax {
public:	T operator()( T a, T b ) { return ( max( a, b ) ); }
};
template<typename T = ll> class CompareMin {
public:	T operator()( T a, T b ) { return ( min( a, b ) ); }
};

template<typename T = ll, typename F = CompareMax<T>, T I = 0> class SegTree {
public:
	SegTree( T n ) { N = n; v.resize( 4 * n ); rep( i, 4 * n ) v[i] = I; }
	void update( T i, T x ) { i += N - 1; v[i] = x; while ( i > 0 ) {
			i = ( i - 1 ) / 2; v[i] = func( v[i * 2 + 1], v[i * 2 + 2] ); } }
	T query( T a, T b, T k, T l, T r ) {
		if ( r <= a || b <= l ) return ( I );
		if ( a <= l && r <= b ) return ( v[k] );
		else { T vl = query( a, b, k * 2 + 1, l, ( l + r ) / 2 );
			T vr = query( a, b, k * 2 + 2, ( l + r ) / 2, r );
			return ( func( vl, vr ) ); }
	}
private:
	T N; F func; vector<T> v;
};

template<typename T = ll> T solveLIS( const vector<T> &v ) {
	vector<T> dp( v.size(), numeric_limits<T>::max() );
	rep( i, v.size() ) {
		*lower_bound( ALL( dp ), v[i] ) = v[i];
	}
	return ( distance( dp.begin(), lower_bound( ALL( dp ), numeric_limits<T>::max() ) ) );
}

template<typename T = ll> T power( T a, T b, T m = MOD ) {
	T res = 1;
	while ( b > 0 ) { if ( b & 1 ) res = res * a % m;
		a = a * a % m; b >>= 1; }
	return ( res );
}

/*
constexpr ll COMBSIZE = 500010;
ll fact[COMBSIZE];
ll inv[COMBSIZE];
ll inv_fact[COMBSIZE];

template<typename T> void initComb( T n, T m = MOD )
{
	fact[0] = fact[1] = inv_fact[0] = inv_fact[1] = 1;
	inv[1] = 1;
	for ( int i = 2; i < n; i++ ) {
		fact[i] = ( fact[i - 1] * i ) % m;
		inv[i] = m - inv[m % i] * ( m / i ) % m;
		inv_fact[i] = inv_fact[i - 1] * inv[i] % m;
	}
}

template<typename T> T comb( T n, T r, T m = MOD )
{
	if ( n < r ) return ( 0 );
	if ( n < 0 || r < 0 ) return ( 0 );
	return ( fact[n] * ( inv_fact[r] * inv_fact[n - r] % m ) % m );
}
*/

/*
	regex reg( R"(^(dream|dreamer|erase|eraser)+$)" );
	smatch m;

	if ( regex_match( s, m, reg ) )
	{
		cout << "YES" << endl;
	}
	else
	{
		cout << "NO" << endl;
	}
*/
/*
	cout << setprecision( 16 );
	cout << fixed << setprecision( 16 );
*/

void replace( string &s, string t, string r ) {
	string::size_type p = 0;
	while ( ( p = s.find( t, p ) ) != string::npos ) {
		s.replace( p, t.length(), r );
		p += r.length();
	}
}


int main()
{
	ll N, x;
	cin >> N >> x;

	ll cnt = 0;
	vector<ll> v( N );
	rep( i, N )
	{
		cin >> v[i];
		if ( v[i] > x )
		{
			cnt += v[i] - x;
			v[i] = x;
		}
	}

	rep( i, N )
	{
		if ( i % 2 )
		{
			if ( v[i - 1] + v[i] > x )
			{
				ll tmp = v[i - 1] + v[i] - x;
				cnt += tmp;
				v[i] -= tmp;
			}
		}
	}

	cout << cnt << endl;

	return ( 0 );
}