CodeForces978E Bus Video System（模拟&细节）

The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.

If xx is the number of passengers in a bus just before the current bus stop and yy is the number of passengers in the bus just after current bus stop, the system records the number y−xy−x. So the system records show how number of passengers changed.

The test run was made for single bus and nn bus stops. Thus, the system recorded the sequence of integers a1,a2,…,ana1,a2,…,an (exactly one number for each bus stop), where aiaiis the record for the bus stop ii. The bus stops are numbered from 11 to nn in chronological order.

Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to ww (that is, at any time in the bus there should be from 00 to ww passengers inclusive).

Input

The first line contains two integers nn and ww (1≤n≤1000,1≤w≤109)(1≤n≤1000,1≤w≤109) — the number of bus stops and the capacity of the bus.

The second line contains a sequence a1,a2,…,ana1,a2,…,an (−106≤ai≤106)(−106≤ai≤106), where aiai equals to the number, which has been recorded by the video system after the ii-th bus stop.

Output

Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to ww. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.

Examples

Input

3 5
2 1 -3

Output

3

Input

2 4
-1 1

Output

4

Input

4 10
2 4 1 2

Output

2

Note

In the first example initially in the bus could be 00, 11 or 22 passengers.

In the second example initially in the bus could be 11, 22, 33 or 44 passengers.

In the third example initially in the bus could be 00 or 11 passenger.

题意：给出n和w，n表示n站，w表示公交车的最大容量。

给出每一站公交车的上下人情况，正数表示上人，负数表示下人，问在一开始的时候公交车有多少种可能的人数。

解题思路：求出车中上人的数量的最大值maxx，和下人的最大值minn。

如果每一站全是上人，那么初始时的人数为w-maxx+1。

如果每一站都是下人，那么初始时的人数为w+minn+1

如果既有上人也有下人，那么初始人数为最小值到最大值区间里的人数，如果这个人数不符合条件则输出0。

AC代码：

#include<stdio.h>
#define INF 0x3f3f3f3f

int main()
{
	int n,m,x;
	while(~scanf("%d%d",&n,&m))
	{
		int sum=0,minn=INF,maxx=-INF;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			sum+=x;
			if(sum>maxx) maxx=sum;
			if(sum<minn) minn=sum;
		}
		
		if(-minn>m||maxx>m) printf("0\n");//超出车的容量 
		else if(minn>=0) printf("%d\n",m-maxx+1);//每一站全是上人 
		else if(maxx<=0) printf("%d\n",m+minn+1);//每一站全是下人 
		else//有上人也有下人 
		{
			int s1=-minn;
			int s2=m-maxx; 
			if(s2-s1+1>0&&s2-s1+1<=m) printf("%d\n",s2-s1+1);
			else printf("0\n");
		}
	}
	return 0;
}