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

int main() {
    string s;
    cin >> s;
    int n = s.length();
    if(s[0] == '0' || s[n - 1] == '1'){
        cout << -1;
        // cout << "666";
        return 0;
    }

    for(int i = 0; i < n - 1; i++) {
        if(s[i] == '1' && s[n - 1 - i - 1] == '0') {
            cout << -1;
            // cout << 777;
            return 0;
        }
    }

    int now = 0;
    for(int i = 1; i < n; i++) {
        printf("%d %d\n", now + 1, i + 1);
        if(s[i - 1] == '1') now = i;
    }
}
### 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;

int main() {
    string s;
    cin >> s;
    int n = s.length();
    if(s[0] == '0' || s[n - 1] == '1'){
        cout << -1;
        // cout << "666";
        return 0;
    }

    for(int i = 0; i < n - 1; i++) {
        if(s[i] == '1' && s[n - 1 - i - 1] == '0') {
            cout << -1;
            // cout << 777;
            return 0;
        }
    }

    int now = 0;
    for(int i = 1; i < n; i++) {
        printf("%d %d\n", now + 1, i + 1);
        if(s[i - 1] == '1') now = i;
    }
}
```