#include <stdio.h>

#define MAXN 34569

int N, N2, T;
int x[MAXN], y[MAXN], d[MAXN], s[MAXN];
long long best[MAXN];

int main()
{
	freopen("euro.in", "rt", stdin);
	freopen("euro.out", "wt", stdout);
	scanf("%d %d", &N, &T);
	int i, S = 0;
	s[ N2 = 0 ] = 0;
	for (i = 1; i <= N; i++)
	{
		scanf("%d", x + i);
		S += x[i];			//atata timp cat am in cont o valoare pozitiva nu voi obtine profitul maxim daca fac conversie
		if (S < 0)			//abia atunci cand obtin in cont o valoare negativa trebuie sa iau in considerare conversia
		{				//=> comprim intervalele cu valori pozitive
			y[ ++N2 ] = S;
			s[ N2 ] = s[ N2 - 1 ] + S;
			d[ N2 ] = i;
			S = 0;
		}
	}
	if (d[ N2 ] != N)
	{
		y[ ++N2 ] = S;
		s[ N2 ] = s[ N2 - 1 ] + S;
		d[ N2 ] = N;
	}

	best[0] = 0;
	for (i = 1; i <= N2; i++)
	{
		best[i] = -(1LL << 60);

		int j; long long S2 = 0;
		for (j = 1; j <= i; j++)
		{
			S2 = (long long)d[i] * (s[i] - s[j - 1]);

			if (best[j - 1] + S2 - T > best[i])
				best[i] = best[j - 1] + S2 - T;
		}
	}
	
	printf("%lld\n", best[ N2 ]);
	return 0;
}