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

char str[100001];

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

### 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;

char str[100001];

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

```