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.
## necessary imports
import sys
input = sys.stdin.readline
#from math import ceil, floor, factorial;

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return int(res) 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

# old code 
# n = int(input());
# ans = '';
# while(n > 0):
#     n -= 1;
#     remainder = n % 26;
#     digit = chr(remainder + 97);
#     ans = digit + ans;
#     n = (n - remainder) // 26;
#     # print(n, remainder, digit, ans);
# print(ans);

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

## PROBLEM A
# for _ in range(int(input())):
#     n = int(input());
#     if n % 4 == 0:
#         print('YES');
#     else:
#         print('NO');

## PROBLEM B
for _ in range(int(input())):
    n = int(input()); s = list(input().strip());
    start = int(s[0]); counts = [];
    this = 0; prev = None;
    for i in range(n):
        if prev == None:
            prev = s[i]; this += 1;
        else:
            if prev == s[i]:
                this += 1;
            else:
                counts.append(this);
                this = 1;
            prev = s[i];
        if i == n-1:
            counts.append(this);
    ans1 = []; ans2 = [];
    if start == 0:
        ans1 = ['0']*counts[0];
        if len(counts) % 2 == 0:
            ans2 = ['1']*counts[-1];

            if len(counts) > 2:
                ans = ans1 + ['0'] + ans2;
            else:
                ans = ans1 + ans2;
        else:
            if len(counts) > 1:
                ans = ans1 + ['0'] + ans2;
            else:
                ans = ans1 + ans2;        
    else:
        if len(counts) & 1:
            if len(counts) == 1:
                ans = ['1']*counts[-1];
            else:
                ans = ['0'] + ['1']*counts[-1];
        else:
            ans = ['0'];
    ans = ''.join(ans);
    print(ans);