#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

const int MAXN = 1e5 + 7;
char s[MAXN];

int main() {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    if (s[1] == '0' || s[len] == '1') return printf("-1\n"), 0;
    for (int i = 1; i <= len / 2; i++) if (s[i] ^ s[len - i]) return printf("-1\n"), 0;
    int x = 1;
    for (int i = 2; i <= len; i++) {
        printf("%d %d\n", x, i);
        if (s[i - 1] == '1') x = 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
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

const int MAXN = 1e5 + 7;
char s[MAXN];

int main() {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    if (s[1] == '0' || s[len] == '1') return printf("-1\n"), 0;
    for (int i = 1; i <= len / 2; i++) if (s[i] ^ s[len - i]) return printf("-1\n"), 0;
    int x = 1;
    for (int i = 2; i <= len; i++) {
        printf("%d %d\n", x, i);
        if (s[i - 1] == '1') x = i; 
    }
    return 0;
}
```