#include <bits/stdc++.h>

using namespace std;

char s[100010];

int main()
{
    s[0] = '0';
    scanf("%s", s + 1);
    int n = strlen(s);
    int i = 0, j = n - 1;
    if(s[1] != '1')
    {
        printf("-1");
        return 0;
    }
    for(; i < j; i++, j--)
    {
        if(s[i] != s[j])
        {
            printf("-1");
            return 0;
        }
    }
    int ult = 1;
    for(i = 1; i < n - 1; i++)
    {
        printf("%d %d\n", ult, i + 1);
        if(s[i] == '1')
            ult = i + 1;
    }
    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 <bits/stdc++.h>

using namespace std;

char s[100010];

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

```