#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; ++i)

int main() {
  string s;
  cin >> s;
  int n = s.size(), u = 1, v = 1;
  bool b = false;
  rep(i, n / 2) if(s[i] != s[n - 2 - i]) b = true;
  if(b || s[0] != '1' || s[n - 1] != '0') {
    cout << -1 << endl;
    return 0;
  }
  rep(i, n / 2 + 1) s[n / 2 + 1 + i] = '0';
  rep(i, n - 1) {
    cout << u << ' ' << ++v << '\n';
    if(s[n - 2 - i] == '1') u = v;
  }
}
### 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;
#define rep(i, n) for(int i = 0; i < n; ++i)

int main() {
  string s;
  cin >> s;
  int n = s.size(), u = 1, v = 1;
  bool b = false;
  rep(i, n / 2) if(s[i] != s[n - 2 - i]) b = true;
  if(b || s[0] != '1' || s[n - 1] != '0') {
    cout << -1 << endl;
    return 0;
  }
  rep(i, n / 2 + 1) s[n / 2 + 1 + i] = '0';
  rep(i, n - 1) {
    cout << u << ' ' << ++v << '\n';
    if(s[n - 2 - i] == '1') u = v;
  }
}
```