There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.

At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?

Constraints

* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the minimum possible number of different characters contained in the strings.

Examples

Input

3
3 2 1


Output

2


Input

5
2 3 2 1 2


Output

2
def main():
    import heapq
    n = int(input())
    a = list(map(int, input().split()))
    for i in range(n-1):
        if a[i] >= a[i+1]:
            break
    else:
        print(1)
        return

    h = []
    heapq.heapify(h)
    kind = 1
    length = 0
    d = dict()
    for i in a:
        if length >= i:
            while h:
                hh = -heapq.heappop(h)
                if hh > i:
                    d.pop(hh)
                else:
                    heapq.heappush(h, -hh)
                    break
            for j in range(i, 0, -1):
                if j not in d:
                    d[j] = 2
                    heapq.heappush(h, -j)
                    kind = max(kind, 2)
                    while h:
                        hh = -heapq.heappop(h)
                        if hh > j:
                            d.pop(hh)
                        else:
                            heapq.heappush(h, -hh)
                            break
                    break
                else:
                    if d[j] < kind:
                        d[j] += 1
                        while h:
                            hh = -heapq.heappop(h)
                            if hh > j:
                                d.pop(hh)
                            else:
                                heapq.heappush(h, -hh)
                                break
                        break
            else:
                d[i] += 1
                kind += 1
        length = i

    def value(kind):
        h = []
        heapq.heapify(h)
        length = 0
        d = dict()
        for i in a:
            if length >= i:
                while h:
                    hh = -heapq.heappop(h)
                    if hh > i:
                        d.pop(hh)
                    else:
                        heapq.heappush(h, -hh)
                        break
                for j in range(i, 0, -1):
                    if j not in d:
                        d[j] = 2
                        heapq.heappush(h, -j)
                        while h:
                            hh = -heapq.heappop(h)
                            if hh > j:
                                d.pop(hh)
                            else:
                                heapq.heappush(h, -hh)
                                break
                        break
                    else:
                        if d[j] < kind:
                            d[j] += 1
                            while h:
                                hh = -heapq.heappop(h)
                                if hh > j:
                                    d.pop(hh)
                                else:
                                    heapq.heappush(h, -hh)
                                    break
                            break
                else:
                    return False
            length = i
        return True

    def b_search(ok, ng, value):
        while abs(ok-ng) > 1:
            mid = (ok+ng)//2
            if value(mid):
                ok = mid
            else:
                ng = mid
        return ok

    print(b_search(kind, max(1, kind-30), value))


main()