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&nbs***bsp;22 passengers.

In the second example initially in the bus could be 11, 22, 33&nbs***bsp;44 passengers.

In the third example initially in the bus could be 00&nbs***bsp;11 passenger.

 

题意：

 给你一个含有n个整数的数组，每一个数a[i]代表汽车在站i时，车上增多了a[i]个人，如果a[i]为负，代表减少了人数。

并告诉你这个汽车的最大承载力为w个人，

请你判断初始时汽车上有多少个人，才满足整个数组的情况，。

如果某一个情况，车上的人数为负，或者人数大于w，那么说明这个数组时不合理的，。这时请输出0

思路：

可以抽象为，求这个数组的前缀和数组中的最大值和最小值，。只要最大值不大于容量，再判断下最低值的绝对值不大于容量。就可以说明是合理的。

然后可以的方案数中初始的人数一定是连续的，那么这些人数中的最大值是min(w-maxsum,w) ，即不让过程中容量大于w的最大值。

最小值是max（0,-1*minsum），然后最大值减去最小值+1就是答案了。

 

细节见代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
// HFUU-QerM
// 21:49:59
ll n;
ll w;
ll a[maxn];
int main()
{
    //freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    gbtb;
    cin >> n >> w;
    repd(i, 1, n)
    {
        cin >> a[i];
    }
    ll f = -1e18;
    ll g = 1e18;
    ll v = 0ll;
    repd(i, 1, n)
    {
        v += a[i];
        f = max(f, v);
        g = min(g, v);
    }
    // db(f);
    // db(g);
    if ((abs(f)) > w || abs(g) > w)
    {
        cout << 0 << endl;
    } else
    {
        ll s = w - f;
        ll x = 0ll;
        // db(s);
        s=min(s,w);
        if (g < 0)
        {
            x = -1 * g;
        }
        // db(x);
        if (x > s)
        {
            cout << 0 << endl;
        } else
        {
            cout << s - x + 1ll << endl;
        }
    }




    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}