human
I found an interesting problem on https://codeforces.com/problemset/problem/1436/E:
In this problem MEX of a certain array is the smallest positive integer not contained in this array.

Everyone knows this definition, including Lesha. But Lesha loves MEX, so he comes up with a new problem involving MEX every day, including today.

You are given an array $a$ of length $n$. Lesha considers all the non-empty subarrays of the initial array and computes MEX for each of them. Then Lesha computes MEX of the obtained numbers.

An array $b$ is a subarray of an array $a$, if $b$ can be obtained from $a$ by deletion of several (possible none or all) elements from the beginning and several (possibly none or all) elements from the end. In particular, an array is a subarray of itself.

Lesha understands that the problem is very interesting this time, but he doesn't know how to solve it. Help him and find the MEX of MEXes of all the subarrays!

-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the array. 

The next line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the elements of the array.

-----Output-----

Print a single integer — the MEX of MEXes of all subarrays.

-----Examples-----
Input
3
1 3 2

Output
3

Input
5
1 4 3 1 2

Output
6

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a Python code to find the MEX of MEXes of all the subarrays!?
assistant
#Here is Python code to find the MEX of MEXes of all the subarrays!
.

class SegmentTree:
    def __init__(self, init_val, segfunc, ide_ele):
        n = len(init_val)
        self.segfunc = segfunc
        self.ide_ele = ide_ele
        self.num = 1 << (n - 1).bit_length()
        self.tree = [ide_ele] * 2 * self.num
        self.range = [(-1,n)] * 2 * self.num
        # 配列の値を葉にセット
        for i in range(n):
            self.tree[self.num + i] = init_val[i]
            self.range[self.num + i] = (i,i)
        # 構築していく
        for i in range(self.num - 1, 0, -1):
            self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
            self.range[i] = (self.range[2 * i][0],self.range[2 * i + 1][1])

    def update(self, k, x):
        k += self.num
        self.tree[k] = x
        while k > 1:
            self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
            k >>= 1

    def query(self, l, r):
        res = self.ide_ele

        l += self.num
        r += self.num
        while l < r:
            if l & 1:
                res = self.segfunc(res, self.tree[l])
                l += 1
            if r & 1:
                res = self.segfunc(res, self.tree[r - 1])
            l >>= 1
            r >>= 1
        return res

    def bisect_l(self,l,r,x):
        l += self.num
        r += self.num
        Lmin = -1
        Rmin = -1
        while l<r:
            if l & 1:
                if self.tree[l] <= x and Lmin==-1:
                    Lmin = l
                l += 1
            if r & 1:
                if self.tree[r-1] <=x:
                    Rmin = r-1
            l >>= 1
            r >>= 1

        if Lmin != -1:
            pos = Lmin
            while pos<self.num:
                if self.tree[2 * pos] <=x:
                    pos = 2 * pos
                else:
                    pos = 2 * pos +1
            return pos-self.num
        elif Rmin != -1:
            pos = Rmin
            while pos<self.num:
                if self.tree[2 * pos] <=x:
                    pos = 2 * pos
                else:
                    pos...