#include <stdio.h>

char s[100005];
int n;

int main()
{
    #ifdef NOT_DMOJ
    freopen("data.txt","r",stdin);
    #endif // NOT_DMOJ
    scanf("%s%n",s+1,&n);
    // do a basic check
    if (s[1]=='0' || s[n]=='1') {
        printf("-1\n");
        return 0;
    }
    for (int i=1; i<n; i++) {
        if (s[i]!=s[n-i]) {
            printf("-1\n");
            return 0;
        }
    }
    int parent=1;
    for (int i=2; i<=n; i++) {
        printf("%d %d\n",parent,i);
        if (s[n-i+1]=='1') {
            parent=i;
        }
    }
}

### 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 <stdio.h>

char s[100005];
int n;

int main()
{
    #ifdef NOT_DMOJ
    freopen("data.txt","r",stdin);
    #endif // NOT_DMOJ
    scanf("%s%n",s+1,&n);
    // do a basic check
    if (s[1]=='0' || s[n]=='1') {
        printf("-1\n");
        return 0;
    }
    for (int i=1; i<n; i++) {
        if (s[i]!=s[n-i]) {
            printf("-1\n");
            return 0;
        }
    }
    int parent=1;
    for (int i=2; i<=n; i++) {
        printf("%d %d\n",parent,i);
        if (s[n-i+1]=='1') {
            parent=i;
        }
    }
}

```