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

using ll = long long;

const int N = 100005;

char s[N];
int n;

void F(){ puts("-1"); exit(0); }

int main(){
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for(int i = 1; i <= n; i++) s[i] -= '0';
    if(!s[1] || s[n]) F();
    for(int i = 1; i <= n / 2; i++) if(s[i] != s[n - i]) F();
    puts("1 2");
    for(int i = 2, c = 1; i < n; i++){
        printf("%d %d\n", c, i + 1);
        if(s[i]) c = i + 1;
    }
}
### Prompt

Please create a solution in cpp 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;

using ll = long long;

const int N = 100005;

char s[N];
int n;

void F(){ puts("-1"); exit(0); }

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