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, ..., 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 ≤ n ≤ 1 000, 1 ≤ w ≤ 10^{9}) — the number of bus stops and the capacity of the bus.

The second line contains a sequence a_1, a_2, ..., a_n (-10^{6} ≤ a_i ≤ 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

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 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.
z,zz,dgraphs=input,lambda:list(map(int,z().split())),{}
from string import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
    p=ceil(x**.5)+1
    for i in range(2,p):
        if x%i==0:return 0
    return 1
def dfs(u,visit,graph):
    visit[u]=True
    for i in graph[u]:
        if not visit[i]:
            dfs(i,visit,graph)
################################################################################

"""

led=(6,2,5,5,4,5,6,3,7,6)

vowel={'a':0,'e':0,'i':0,'o':0,'u':0}

color4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]
z,zz,dgraphs=input,lambda:list(map(int,z().split())),{}
d,l,r,b=[],0,1,{}
for _ in range(int(z())):d.append(z().split())
for i in d:
    id_ =int(i[1])
    if i[0]=='L':
        b[id_]=l
        l-=1
    if i[0]=='R':
        b[id_]=r
        r+=1
    if i[0]=='?':
        p = b[id_]
        print(min(p-l-1,r-p-1))
    
    
"""

###########################---START-CODING---####################################
n,w=zz()
l=zz()

mi,mx=0,0
t=0
for i in l:
    t+=i
    mx=max(t,mx)
    mi=min(mi,t)
print(max(0,w-mx+1+mi))
    
    
    
    
    

    
