In the 2022 year, Mike found two binary integers a and b of length n (both of them are written only by digits 0 and 1) that can have leading zeroes. In order not to forget them, he wanted to construct integer d in the following way: 

  * he creates an integer c as a result of bitwise summing of a and b without transferring carry, so c may have one or more 2-s. For example, the result of bitwise summing of 0110 and 1101 is 1211 or the sum of 011000 and 011000 is 022000; 
  * after that Mike replaces equal consecutive digits in c by one digit, thus getting d. In the cases above after this operation, 1211 becomes 121 and 022000 becomes 020 (so, d won't have equal consecutive digits). 



Unfortunately, Mike lost integer a before he could calculate d himself. Now, to cheer him up, you want to find any binary integer a of length n such that d will be maximum possible as integer.

Maximum possible as integer means that 102 > 21, 012 < 101, 021 = 21 and so on.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

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

The second line of each test case contains binary integer b of length n. The integer b consists only of digits 0 and 1.

It is guaranteed that the total sum of n over all t test cases doesn't exceed 10^5.

Output

For each test case output one binary integer a of length n. Note, that a or b may have leading zeroes but must have the same length n.

Example

Input


5
1
0
3
011
3
110
6
111000
6
001011


Output


1
110
100
101101
101110

Note

In the first test case, b = 0 and choosing a = 1 gives d = 1 as a result.

In the second test case, b = 011 so: 

  * if you choose a = 000, c will be equal to 011, so d = 01; 
  * if you choose a = 111, c will be equal to 122, so d = 12; 
  * if you choose a = 010, you'll get d = 021. 
  * If you select a = 110, you'll get d = 121. 

We can show that answer a = 110 is optimal and d = 121 is maximum possible.

In the third test case, b = 110. If you choose a = 100, you'll get d = 210 and it's the maximum possible d.

In the fourth test case, b = 111000. If you choose a = 101101, you'll get d = 212101 and it's maximum possible d.

In the fifth test case, b = 001011. If you choose a = 101110, you'll get d = 102121 and it's maximum possible d.
#include<iostream>
#include<vector>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
#include<queue>
#include<stack>
#include<cstdio>
#define ll long long
using namespace std;

inline int read(){
    int s = 0, w = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-') w = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9'){
        s = s * 10 + c - '0';
        c = getchar();
    }
    return s * w;
}

int main(){
    int t, n;
    int b[1000001], a[1000001];
    t = read();
    while(t--){
        n = read();
        memset(b, 0, n + 1);
        memset(a, 0, n + 1);
        for(int i = 1; i <= n; ++i){
            char c;
            scanf("%c", &c);
            b[i] = c - '0';
        }
        a[1] = 1;
        for(int i = 2; i <= n; ++i){
            if(b[i - 1] + a[i - 1] == 2){
                if(b[i] == 0) a[i] = 1;
                else if(b[i] == 1) a[i] = 0;
            }
            else if(b[i - 1] + a[i - 1] == 1){
                if(b[i] == 0) a[i] = 0;
                else a[i] = 1;
            }
            else if(b[i - 1] + a[i - 1] == 0){
                a[i] = 1;
            }
        }
        for(int i = 1; i <= n; i++) cout << a[i];
        cout << endl;
    }
    return 0;
}

