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.
from sys import *
from collections import *
from math import *
from random import *
from datetime import date

class BIT:
  def __init__(self, n):
  
    self.n = n  
    self.tree = [0]*(n+1)
    
  def __setitem__(self, i, v):

    while i <= self.n:
      self.tree[i] += v
      i += i&-i
      
  def __getitem__(self, i):
    
    p = 0

    while i > 0:
      p += self.tree[i]
      i -= i&-i
    return p
    
def lStr(d = " "): return input().split(" ")
def lInt(d = " "): return [int(i) for i in input().split(d)]
def lDec(d = " "): return [float(i) for i in input().split(d)]

n, *t = lInt()
p = lInt()
b = BIT(n+1)
maxi = 0
rec = [0]*(n+1)
block = [0]*(n+1)
ans, best = 10000000, 0
fix = 0

for i in range(0, n):
  v = p[i]
  b[v] = 1
  g = i-b[v-1]
  rec[v] = g == 0
  fix += g == 0
  block[maxi] += g == 1
  maxi = max(maxi, v)
for v in p:
  tot = fix+block[v]-(rec[v] == 1)

  if tot > best:
    best = tot
    ans = v
  if tot == best:
    ans = min(ans, v)
print(ans)
  


