#include <bits/stdc++.h>
#define REP(i, a, n) for (ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

int main(void) {
  string S;
  cin >> S;
  ll N = S.length();

  bool ok = true;
  ok = ok && S[N - 1] == '0';
  ok = ok && S[N - 2] == '1';
  REP(i, 0, N - 1) ok = ok && S[i] == S[N - 2 - i];
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }

  ll prev = 1, current = 2;
  REP(i, 0, N - 1) {
    cout << prev << " " << current << endl;
    if (S[i] == '1') prev = current;
    current++;
  }
}
### Prompt

Generate 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>
#define REP(i, a, n) for (ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

int main(void) {
  string S;
  cin >> S;
  ll N = S.length();

  bool ok = true;
  ok = ok && S[N - 1] == '0';
  ok = ok && S[N - 2] == '1';
  REP(i, 0, N - 1) ok = ok && S[i] == S[N - 2 - i];
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }

  ll prev = 1, current = 2;
  REP(i, 0, N - 1) {
    cout << prev << " " << current << endl;
    if (S[i] == '1') prev = current;
    current++;
  }
}
```