#include<bits/stdc++.h>
using namespace std;
void check(bool p)
{
  if(!p)
  {
    puts("-1");
    exit(0);
  }
}
string s; int N;
int main()
{
  cin >> s; N = s.size(); s = '0' + s;
  check(s[0] == s[N]);
  vector<int> V;
  for(int i=1; i<=N/2; ++i)
  {
    check(s[i] == s[N-i]);
    if(s[i] == '1') V.push_back(i);
  }
  check(s[1] == '1');
  V.push_back(N);
  reverse(V.begin(), V.end());
  for(int i=0; i<(int)V.size()-1; ++i)
    for(int j=V[i+1]; j<V[i]; ++j) printf("%d %d\n", V[i], j);
  
  return 0;
}
### 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;
void check(bool p)
{
  if(!p)
  {
    puts("-1");
    exit(0);
  }
}
string s; int N;
int main()
{
  cin >> s; N = s.size(); s = '0' + s;
  check(s[0] == s[N]);
  vector<int> V;
  for(int i=1; i<=N/2; ++i)
  {
    check(s[i] == s[N-i]);
    if(s[i] == '1') V.push_back(i);
  }
  check(s[1] == '1');
  V.push_back(N);
  reverse(V.begin(), V.end());
  for(int i=0; i<(int)V.size()-1; ++i)
    for(int j=V[i+1]; j<V[i]; ++j) printf("%d %d\n", V[i], j);
  
  return 0;
}
```