On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you've created a program to compress these graphs.

The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.

More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n-k+1, such that $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$

For example, the 3-compression array of [1, 3, 4, 5, 2] is [min\{1, 3, 4\}, min\{3, 4, 5\}, min\{4, 5, 2\}]=[1, 3, 2].

A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).

A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤ k≤ n if CodeCook users will be happy after a k-compression of this array or not.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the array.

The second line of the description of each test case contains n integers a_1,…,a_n (1≤ a_i≤ n) — the elements of the array.

It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.

Output

For each test case, print a binary string of length n. 

The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise. 

Example

Input


5
5
1 5 3 4 2
4
1 3 2 1
5
1 3 3 3 2
10
1 2 3 4 5 6 7 8 9 10
3
3 3 2


Output


10111
0001
00111
1111111111
000

Note

In the first test case, a=[1, 5, 3, 4, 2].

  * The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation. 
  * The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice. 
  * The 3-compression of a is [1, 3, 2] and it is a permutation. 
  * The 4-compression of a is [1, 2] and it is a permutation. 
  * The 5-compression of a is [1] and it is a permutation. 
import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for k in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10

class BIT:

    def __init__(self, n):
        self.n = n
        # 0-indexed
        n += 1
        nv = 1
        while nv < n:
            nv *= 2
        self.size = nv
        self.tree = [0] * nv

    def sum(self, i):
        s = 0
        i += 1
        while i > 0:
            s += self.tree[i-1]
            i -= i & -i
        return s

    def add(self, i, x):
        i += 1
        while i <= self.size:
            self.tree[i-1] += x
            i += i & -i

    def query(self, l, r):
        return self.sum(r-1) - self.sum(l-1)
    
    def get(self, i):
        return self.query(i, i+1)

    def update(self, i, x):
        self.add(i, x - self.get(i))

    def print(self):
        for i in range(self.n):
            print(self.get(i), end=' ')
        print()

    def bisearch_fore(self, l, r, x):
        l_sm = self.sum(l-1)
        ok = r + 1
        ng = l - 1
        while ng+1 < ok:
            mid = (ok+ng) // 2
            if self.sum(mid) - l_sm >= x:
                ok = mid
            else:
                ng = mid
        if ok != r + 1:
            return ok
        else:
            return -1

    def bisearch_back(self, l, r, x):
        r_sm = self.sum(r)
        ok = l - 1
        ng = r + 1
        while ok+1 < ng:
            mid = (ok+ng) // 2
            if r_sm - self.sum(mid-1) >= x:
                ok = mid
            else:
                ng = mid
        if ok != l - 1:
            return ok
        else:
            return -1

for _ in range(INT()):
    N = INT()
    A = [a-1 for a in LIST()]
    C = [0] * N
    B = [[] for i in range(N)]
    for i in range(N):
        C[A[i]] += 1
        B[A[i]].append(i)

    bit = BIT(N)
    ans = [0] * N
    for a in range(N):
        if not C[a]:
            break
        ln = N - a - 1
        se = set()
        for i in B[a]:
            l = bit.bisearch_back(max(i-ln, 0), i-1, 1)
            if l == -1:
                l = max(i-ln, 0)
            else:
                l += 1
            r = bit.bisearch_fore(i+1, min(i+ln, N-1), 1)
            if r == -1:
                r = min(i+ln, N-1)
            else:
                r -= 1
            if r-l == ln:
                se.add((l, r))
            bit.add(i, 1)
        if len(se) != 1:
            break
        ans[ln] = 1
    if len(set(A)) == N:
        ans[0] = 1
    print(''.join(map(str, (ans))))
