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

#define rep(i,n) for(int i=0,i##_cond=(n);i<i##_cond;++i)
#define FOR(i,a,b) for(int i=(a),i##_cond=(b);i<i##_cond;++i)
#define ROF(i,a,b) for(int i=(a)-1,i##_cond=(b);i>=i##_cond;--i)

int main(){
  string s; cin >> s;
  int n = s.size();

  bool f = false;
  if(s[0] == '0' or s[n-1] == '1') f = true;
  rep(i,n/2)
    if(s[i] != s[n-i-2]) f = true;
  if(f){
    cout << -1 << endl;
    return 0;
  }

  int k = n;
  ROF(i,n-1,0){
    cout << i+1 << " " << k << endl;
    if(s[i] == '1') k = i + 1;
  }
}
### Prompt

Your challenge is to write a Cpp solution to the following problem:

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##_cond=(n);i<i##_cond;++i)
#define FOR(i,a,b) for(int i=(a),i##_cond=(b);i<i##_cond;++i)
#define ROF(i,a,b) for(int i=(a)-1,i##_cond=(b);i>=i##_cond;--i)

int main(){
  string s; cin >> s;
  int n = s.size();

  bool f = false;
  if(s[0] == '0' or s[n-1] == '1') f = true;
  rep(i,n/2)
    if(s[i] != s[n-i-2]) f = true;
  if(f){
    cout << -1 << endl;
    return 0;
  }

  int k = n;
  ROF(i,n-1,0){
    cout << i+1 << " " << k << endl;
    if(s[i] == '1') k = i + 1;
  }
}
```