#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
void err() {
  cout << -1 << '\n';
  exit(0);
}
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  string s;
  cin >> s;
  int n = (int) s.size();
  s = '.' + s;
  for (int i = 1; i <= n; i++) {
    if (i == 1 && s[i] == '0') err();
    if (i == n && s[n] == '1') err();
    if (i < n && s[i] != s[n - i]) err();
  }
  int k = n;
  for (int i = n - 1; i >= 1; i--) {
    cout << k << " " << i << '\n';
    if (s[i] == '1') k = i;
  }
  return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
void err() {
  cout << -1 << '\n';
  exit(0);
}
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  string s;
  cin >> s;
  int n = (int) s.size();
  s = '.' + s;
  for (int i = 1; i <= n; i++) {
    if (i == 1 && s[i] == '0') err();
    if (i == n && s[n] == '1') err();
    if (i < n && s[i] != s[n - i]) err();
  }
  int k = n;
  for (int i = n - 1; i >= 1; i--) {
    cout << k << " " << i << '\n';
    if (s[i] == '1') k = i;
  }
  return 0;
}
```