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())
arr = [int(x) for x in input().split()]
REC = 0
recpos = []
prev = []
posit = [False] * n
for i in range(n):
    if(arr[i] > REC):
        recpos.append(i)
        posit[i] = True
        prev.append(REC)
        REC = arr[i]
N = len(recpos)
# print(recpos)
node = arr[recpos[0]]
MAX = 0
for i in range(N-1):
    can = 0
    currMAX = prev[i]
    for j in range(recpos[i]+1, recpos[i+1]):
        if(arr[j] > currMAX):
            currMAX = arr[j]
            can += 1
    if(can == MAX):
        node = min(arr[recpos[i]], node)
        MAX = can
    elif(can > MAX):
        node = arr[recpos[i]]
        MAX = can

currMAX = prev[-1]
can = 0

for j in range(recpos[-1]+1, n):
    if(arr[j] > currMAX):
        currMAX = arr[j]
        can += 1

if(can == MAX):
    node = min(arr[recpos[-1]], node)
    MAX = can
elif(can > MAX):
    node = arr[recpos[-1]]
    MAX = can
if(MAX <= 1):
    ans = n+1
    for i in range(n):
        if(posit[i] == False):
            ans = min(ans, arr[i])
    if(ans == n+1):
        print(1)
    else:
        print(ans)
else:
    print(node)
