#include <bits/stdc++.h>
using namespace std;

void fail(){
    cout << -1 << endl;
    exit(0);
}

int main(){
    int i, j, k;
    string S;
    cin >> S;
    int N = S.size();
    S = ' ' + S;
    
    for(i=1; i<N; i++){
        if(S[i] != S[N-i]) fail();
    }
    if(S[1] == '0' || S[N] == '1') fail();

    int top = 1;
    for(i=1; i<N; i++){
        printf("%d %d\n", top, i+1);
        if(S[i] == '1') top = i+1;
    }
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

You are given a string s of length n. Does a tree with n vertices that satisfies the following conditions exist?

* The vertices are numbered 1,2,..., n.
* The edges are numbered 1,2,..., n-1, and Edge i connects Vertex u_i and v_i.
* If the i-th character in s is `1`, we can have a connected component of size i by removing one edge from the tree.
* If the i-th character in s is `0`, we cannot have a connected component of size i by removing any one edge from the tree.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq n \leq 10^5
* s is a string of length n consisting of `0` and `1`.

Input

Input is given from Standard Input in the following format:


s


Output

If a tree with n vertices that satisfies the conditions does not exist, print `-1`.

If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.

Examples

Input

1111


Output

-1


Input

1110


Output

1 2
2 3
3 4


Input

1010


Output

1 2
1 3
1 4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

void fail(){
    cout << -1 << endl;
    exit(0);
}

int main(){
    int i, j, k;
    string S;
    cin >> S;
    int N = S.size();
    S = ' ' + S;
    
    for(i=1; i<N; i++){
        if(S[i] != S[N-i]) fail();
    }
    if(S[1] == '0' || S[N] == '1') fail();

    int top = 1;
    for(i=1; i<N; i++){
        printf("%d %d\n", top, i+1);
        if(S[i] == '1') top = i+1;
    }
    return 0;
}
```