808_D. Array Division
Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).

Inserting an element in the same position he was erased from is also considered moving.

Can Vasya divide the array after choosing the right element to move and its new position?

Input

The first line contains single integer n (1 ≤ n ≤ 100000) — the size of the array.

The second line contains n integers a1, a2... an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

Examples

Input

3
1 3 2


Output

YES


Input

5
1 2 3 4 5


Output

NO


Input

5
2 2 3 4 5


Output

YES

Note

In the first example Vasya can move the second element to the end of the array.

In the second example no move can make the division possible.

In the third example Vasya can move the fourth element by one position to the left.
```python
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
ans=0
b=[]

    
def hnbhai(tc):
    n=sb()
    a=sd()
    s=sum(a)
    if s%2!=0 or n==1:
        print("NO")
        return
    s//=2
    for t in range(2):
        d={}
        tot=0
        for i in a:
            d[i]=True
            tot+=i
            if d.get(tot-s):
                print("YES")
                return
        a.reverse()
    print("NO")
for _ in range(1):
    hnbhai(_+1)

```