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=input()
nums=map(int,raw_input().split())
i=0;s=set();ans=[-1,0];mx=0 #last largest num before shader
while i<(n-1): #max i is second to last
    if nums[i]>nums[i+1]: #shader found
        shd=nums[i];i+=1;inc=-1
        while i<(n-1) and nums[i]<shd:
            s.add(nums[i])
            if nums[i]>mx:
                inc+=1
                mx=nums[i]
            mx=max(nums[i],mx);i+=1
        if ans[1]<inc: #check for if n=2,ans[1]=0
            ans[1]=inc;ans[0]=shd
        if i==(n-1):
            if nums[i]<shd:
                s.add(nums[i])
                if nums[i]>mx and ans[1]<(inc+1):
                    ans[1]=(inc+1);ans[0]=shd
        mx=shd
    else:
        mx=max(nums[i],mx);i+=1
#processed all elements
if ans[0]==-1:
    if n==1:
        print 1
    elif nums==sorted(nums):
        print 1
    else:
        print min(s)
else:
    print ans[0]
