Lee was cleaning his house for the party when he found a messy string under the carpets. Now he'd like to make it clean accurately and in a stylish way...

The string s he found is a binary string of length n (i. e. string consists only of 0-s and 1-s).

In one move he can choose two consecutive characters s_i and s_{i+1}, and if s_i is 1 and s_{i + 1} is 0, he can erase exactly one of them (he can choose which one to erase but he can't erase both characters simultaneously). The string shrinks after erasing.

Lee can make an arbitrary number of moves (possibly zero) and he'd like to make the string s as clean as possible. He thinks for two different strings x and y, the shorter string is cleaner, and if they are the same length, then the lexicographically smaller string is cleaner.

Now you should answer t test cases: for the i-th test case, print the cleanest possible string that Lee can get by doing some number of moves.

Small reminder: if we have two strings x and y of the same length then x is lexicographically smaller than y if there is a position i such that x_1 = y_1, x_2 = y_2,..., x_{i - 1} = y_{i - 1} and x_i < y_i.

Input

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

Next 2t lines contain test cases — one per two lines.

The first line of each test case contains the integer n (1 ≤ n ≤ 10^5) — the length of the string s.

The second line contains the binary string s. The string s is a string of length n which consists only of zeroes and ones.

It's guaranteed that sum of n over test cases doesn't exceed 10^5.

Output

Print t answers — one per test case.

The answer to the i-th test case is the cleanest string Lee can get after doing some number of moves (possibly zero).

Example

Input


5
10
0001111111
4
0101
8
11001101
10
1110000000
1
1


Output


0001111111
001
01
0
1

Note

In the first test case, Lee can't perform any moves.

In the second test case, Lee should erase s_2.

In the third test case, Lee can make moves, for example, in the following order: 11001101 → 1100101 → 110101 → 10101 → 1101 → 101 → 01.
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *

# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7

# print(math.factorial(5))
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input


def out(var): sys.stdout.write(str(var))  # for fast output, always take string


def lis(): return list(map(int, inp().split()))


def stringlis(): return list(map(str, inp().split()))


def sep(): return map(int, inp().split())


def strsep(): return map(str, inp().split())


def fsep(): return map(float, inp().split())


def inpu(): return int(inp())


# -----------------------------------------------------------------

def regularbracket(t):
    p = 0
    for i in t:
        if i == "(":
            p += 1
        else:
            p -= 1
        if p < 0:
            return False
    else:
        if p > 0:
            return False
        else:
            return True


# -------------------------------------------------
def binarySearchCount(arr, n, key):
    left = 0
    right = n - 1

    count = 0

    while (left <= right):
        mid = int((right + left) / 2)

        # Check if middle element is
        # less than or equal to key
        if (arr[mid] <= key):
            count = mid + 1
            left = mid + 1
        # If key is smaller, ignore right half
        else:
            right = mid - 1
    return count


# ------------------------------reverse string(pallindrome)
def reverse1(string):
    pp = ""
    for i in string[::-1]:
        pp += i
    if pp == string:
        return True
    return False


# --------------------------------reverse list(paindrome)
def reverse2(list1):
    l = []
    for i in list1[::-1]:
        l.append(i)
    if l == list1:
        return True
    return False


def mex(list1):
    # list1 = sorted(list1)
    p = max(list1) + 1
    for i in range(len(list1)):
        if list1[i] != i:
            p = i
            break
    return p


def sumofdigits(n):
    n = str(n)
    s1 = 0
    for i in n:
        s1 += int(i)
    return s1


def perfect_square(n):
    s = math.sqrt(n)
    if s == int(s):
        return True
    return False


# -----------------------------roman
def roman_number(x):
    if x > 15999:
        return
    value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
    symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
    roman = ""
    i = 0
    while x > 0:
        div = x // value[i]
        x = x % value[i]
        while div:
            roman += symbol[i]
            div -= 1
        i += 1
    return roman


def soretd(s):
    for i in range(1, len(s)):
        if s[i - 1] > s[i]:
            return False
    return True


# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
    ct = 0
    for i in range(2, h + 1, 2):
        for j in range(2, w + 1, 2):
            ct += (h - i + 1) * (w - j + 1)
    return ct


def countrhombi2(h, w):
    return ((h * h) // 4) * ((w * w) // 4)


# ---------------------------------
def binpow(a, b):
    if b == 0:
        return 1
    else:
        res = binpow(a, b // 2)
    if b % 2 != 0:
        return res * res * a
    else:
        return res * res


# -------------------------------------------------------
def binpowmodulus(a, b, m):
    a %= m
    res = 1
    while (b > 0):
        if (b & 1):
            res = res * a % m
        a = a * a % m
        b >>= 1
    return res


# -------------------------------------------------------------
def coprime_to_n(n):
    result = n
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            while (n % i == 0):
                n //= i
            result -= result // i
        i += 1
    if (n > 1):
        result -= result // n
    return result


# -------------------prime
def prime(x):
    if x == 1:
        return False
    else:
        for i in range(2, int(math.sqrt(x)) + 1):
            # print(x)
            if (x % i == 0):
                return False
        else:
            return True

def luckynumwithequalnumberoffourandseven(x,n,a):
    if x >= n and str(x).count("4") == str(x).count("7"):
        a.append(x)
    else:
        if x < 1e12:
            luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
            luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
    return a
#----------------------
def luckynum(x,l,r,a):
    if x>=l and x<=r:
        a.append(x)
    if x>r:
        a.append(x)
        return a
    if x < 1e10:
            luckynum(x * 10 + 4, l,r,a)
            luckynum(x * 10 + 7, l,r,a)
    return a

def luckynuber(x, n, a):
    p = set(str(x))
    if len(p) <= 2:
        a.append(x)
    if x < n:
        luckynuber(x + 1, n, a)
    return a


# ------------------------------------------------------interactive problems

def interact(type, x):
    if type == "r":
        inp = input()
        return inp.strip()
    else:
        print(x, flush=True)


# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
    # Initialize result
    count = 0

    # Keep dividing n by
    # 5 & update Count
    while (n >= 5):
        n //= 5
        count += n

    return count


# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
    if len(arr) > 1:

        # Finding the mid of the array
        mid = len(arr) // 2

        # Dividing the array elements
        L = arr[:mid]

        # into 2 halves
        R = arr[mid:]

        # Sorting the first half
        mergeSort(L)

        # Sorting the second half
        mergeSort(R)

        i = j = k = 0

        # Copy data to temp arrays L[] and R[]
        while i < len(L) and j < len(R):
            if L[i] < R[j]:
                arr[k] = L[i]
                i += 1
            else:
                arr[k] = R[j]
                j += 1
            k += 1

        # Checking if any element was left
        while i < len(L):
            arr[k] = L[i]
            i += 1
            k += 1

        while j < len(R):
            arr[k] = R[j]
            j += 1
            k += 1


# -----------------------------------------------lucky number with two lucky any digits
res = set()


def solven(p, l, a, b, n):  # given number
    if p > n or l > 10:
        return
    if p > 0:
        res.add(p)
    solven(p * 10 + a, l + 1, a, b, n)
    solven(p * 10 + b, l + 1, a, b, n)


# problem
"""
n = int(input())
for a in range(0, 10):
    for b in range(0, a):
        solve(0, 0)
print(len(res))
"""


# Python3 program to find all subsets
# by backtracking.

# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
    print(*subset)
    s = sum(subset)
    d.append(s)
    for i in range(index, len(A)):
        # include the A[i] in subset.
        subset.append(A[i])

        # move onto the next element.
        subsetsUtil(A, subset, i + 1, d)

        # exclude the A[i] from subset and
        # triggers backtracking.
        subset.pop(-1)
    return d


def subsetSums(arr, l, r, d, sum=0):
    if l > r:
        d.append(sum)
        return
    subsetSums(arr, l + 1, r, d, sum + arr[l])

    # Subset excluding arr[l]
    subsetSums(arr, l + 1, r, d, sum)
    return d


def print_factors(x):
    factors = []
    for i in range(1, x + 1):
        if x % i == 0:
            factors.append(i)
    return (factors)


# -----------------------------------------------
def calc(X, d, ans, D):
    # print(X,d)
    if len(X) == 0:
        return

    i = X.index(max(X))
    ans[D[max(X)]] = d

    Y = X[:i]
    Z = X[i + 1:]

    calc(Y, d + 1, ans, D)
    calc(Z, d + 1, ans, D)


# ---------------------------------------

def factorization(n, l):
    c = n
    if prime(n) == True:
        l.append(n)
        return l
    for i in range(2, c):
        if n == 1:
            break
        while n % i == 0:
            l.append(i)
            n = n // i
    return l


# endregion------------------------------


def good(b):
    l = []
    i = 0
    while (len(b) != 0):
        if b[i] < b[len(b) - 1 - i]:
            l.append(b[i])
            b.remove(b[i])
        else:
            l.append(b[len(b) - 1 - i])
            b.remove(b[len(b) - 1 - i])
    if l == sorted(l):
        # print(l)
        return True
    return False


# arr=[]
# print(good(arr))
def generate(st, s):
    if len(s) == 0:
        return

    # If current string is not already present.
    if s not in st:
        st.add(s)

        # Traverse current string, one by one
        # remove every character and recur.
        for i in range(len(s)):
            t = list(s).copy()
            t.remove(s[i])
            t = ''.join(t)
            generate(st, t)
    return
#=--------------------------------------------longest increasing subsequence

def largestincreasingsubsequence(A):
    l = [1]*len(A)
    sub=[]
    for i in range(1,len(l)):
        for k in range(i):
            if A[k]<A[i]:
                sub.append(l[k])
        l[i]=1+max(sub,default=0)
    return max(l,default=0)

#----------------------------------longest palindromic substring




# Python3 program for the
# above approach

# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
    # Stores the prefix
    # sum of nums[]
    prefix_sum = 0

    # Stores the bitwise OR of
    # sum of each subsequence
    result = 0

    # Iterate through array nums[]
    for i in range(N):
        # Bits set in nums[i] are
        # also set in result
        result |= nums[i]

        # Calculate prefix_sum
        prefix_sum += nums[i]

        # Bits set in prefix_sum
        # are also set in result
        result |= prefix_sum

    # Return the result
    return result

#l=[]
def OR(a, n):
    ans = a[0]
    for i in range(1, n):
        ans |= a[i]
        #l.append(ans)
    return ans
#print(prime(12345678987766))
"""
def main():
    for _ in range(inpu()):
        n=inpu()
        arr = lis()
        e,o=[],[]
        for i in range(n*2):
            if arr[i]&1:
                o+=[i]
            else:
                e+=[i]
        for i in range(n-1):
            if len(o)>1:
                print(o.pop()+1,o.pop()+1)
            elif len(e)>1:
                print(e.pop()+1,e.pop()+1)
if __name__ == '__main__':
    main()
"""
"""
def main():
    for _ in range(inpu()):
        n = inpu()
        c = 2
        p = []
        while len(p) < 2 and c * c < n:
            if n % c == 0:
                n = n // c
                p.append(c)
            c += 1
        if len(p) == 2 and n not in p:
            print("YES")
            print(*p, n)
        else:
            print("NO")
if __name__ == '__main__':
    main()
"""

def main():
    for _ in range(inpu()):
        n=inpu()
        s = inp()
        one = s.find('1')
        zero = s.rfind('0')
        if zero < one:
            print(s)
        else:
            print('0' + s[0:one] + s[zero + 1:])
if __name__ == '__main__':
    main()
