E. Bus Video System（找刚开始车上，满足的人数，好题，细节多）

E. Bus Video System

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $x$ is the number of passengers in a bus just before the current bus stop and $y$ is the number of passengers in the bus just after current bus stop, the system records the number $y-x$. So the system records show how number of passengers changed.

The test run was made for single bus and $n$ bus stops. Thus, the system recorded the sequence of integers $a_1,a_2,\dots ,a_n$ (exactly one number for each bus stop), where $a_i$ is the record for the bus stop $i$. The bus stops are numbered from $1$ to $n$ 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 $w$ (that is, at any time in the bus there should be from $0$ to $w$ passengers inclusive).

Input

The first line contains two integers $n$ and $w$ $(1\le n\le 1000,1\le w\le {10}^9)$ — the number of bus stops and the capacity of the bus.

The second line contains a sequence $a_1,a_2,\dots ,a_n$ $(-{10}^6\le a_i\le {10}^6)$, where $a_i$ equals to the number, which has been recorded by the video system after the $i$-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 $w$. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.

Examples

input

Copy

3 5
2 1 -3

output

Copy

3

input

Copy

2 4
-1 1

output

Copy

4

input

Copy

4 10
2 4 1 2

output

Copy

2

Note

In the first example initially in the bus could be $0$, $1$ or $2$ passengers.

In the second example initially in the bus could be $1$, $2$, $3$ or $4$ passengers.

In the third example initially in the bus could be $0$ or $1$ passenger.

题意：输入 n，m，n为有n站，m为公交车的容量，下面一行有n个数，为n站上下车的情况，看看刚开始车上可以有多少种情况，满足 给出 n 站的数据，要是没有输出0，要是有，输出 多少种情况；

思路：这道题细节多，找出车上的上限和下限，在看看有没有交叉的情况，一定要注意交叉的情况；

代码中有解释：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define Max 100010
#define INF 0x3f3f3f3f
#define ll long long 
ll a[Max];
int main()
{
	int i,j,n,m;
	while(~scanf("%d%d",&n,&m))
	{
		int x;
		int ma = -INF;
		int sum = 0;
		int flag = 1;
		int mi = INF;
		for(i = 0;i<n;i++)
		{
			scanf("%d",&x);
			sum += x;
			ma = max(ma,sum);   
			mi = min(mi,sum);	// 
		}
		// ma  找出来最大，当 ma为非负时，刚开始在车上的人数，上限不能超过 m - ma;
		// mi 找出最小，当 mi 为非正时，更开始在车上的人数，下限不能低于 m-（-mi）；  
		// 当 上限和下限 有超过两个交叉的话，那么 刚开始车上的人数，没有满足的，输出0；
		// 有一个数字交叉的话，那么就只有一个数字满足，就是交叉的数字，输出1；
		// 当没有交叉时，直接求有几个数字满足就行了；
		// 一定要注意交叉的情况； 
		if(mi < -m||ma > m)
			flag = 0;
		if(mi<0&&ma>0&&-mi+ma>m)
		{
			flag = 0;
		}
		if(!flag)
		{
			printf("0\n");
			continue;
		}
		int k = m;
		if(mi<=0)
		{
			k = m + mi + 1;
		}
		if(ma>=0&&mi<=0)
		{
			k = k - ma;
		}
		else if(ma>=0)
		{
			k = k - ma + 1;
		}
		printf("%d\n",k);
	}
	return 0;
}