You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.

We remind that in a sequence of numbers a1, a2, ..., ak the element ai is a record if for every integer j (1 ≤ j < i) the following holds: aj < ai. 

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the length of the permutation.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation. All the integers are distinct.

Output

Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples

Input

1
1


Output

1


Input

5
5 1 2 3 4


Output

5

Note

In the first example the only element can be removed.
n=int(input().strip())
a=list(map(int,input().strip().split()))
isit=[0]*(n+1)
isit[a[0]]=1
freq=[0]*(n+1)
m1=[0]*n
m2=[0]*n
m1[0]=a[0]
for i in range (1,n):
    if m1[i-1]<a[i]:
        m1[i]=a[i]
        m2[i]=m1[i-1]
    else:
        m1[i]=m1[i-1]
        if m2[i-1]<a[i]:
            m2[i]=a[i]
        else:
            m2[i]=m2[i-1]
tot=1
for i in range(1,n):
    if m1[i]>a[i]:
        if m2[i]==a[i]:
            freq[m1[i]]+=1
    else:
        tot+=1
        isit[a[i]]=1
mx=0
val=a[0]


for i in range(0,n):
    if mx<tot+freq[a[i]]-isit[a[i]]:
        val=a[i]
        mx=tot+freq[a[i]]-isit[a[i]]
    elif mx==tot+freq[a[i]]-isit[a[i]]:
        val=min(val,a[i])

print(val)
        
