//Problem E
#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int n,a[N];
char ch[N];
#define NO {printf("-1");return 0;}
int main(){
    scanf("%s",ch+1);n=strlen(ch+1);
    for(int i=1;i<=n;i++)a[i]=ch[i]^48;
    if(a[n]||!a[1])NO
    for(int i=1;i<n;i++)if(a[i]!=a[n-i])NO
    a[n]=1;int lust=1;
    for(int i=2;i<=n;i++)if(a[i]){
        for(int j=lust+1;j<=i;j++)printf("%d %d\n",lust,j);
        if(i<n)printf("%d %d\n",lust,i+1),lust=i+1;
        
    }
    return 0;
}
### 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
//Problem E
#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int n,a[N];
char ch[N];
#define NO {printf("-1");return 0;}
int main(){
    scanf("%s",ch+1);n=strlen(ch+1);
    for(int i=1;i<=n;i++)a[i]=ch[i]^48;
    if(a[n]||!a[1])NO
    for(int i=1;i<n;i++)if(a[i]!=a[n-i])NO
    a[n]=1;int lust=1;
    for(int i=2;i<=n;i++)if(a[i]){
        for(int j=lust+1;j<=i;j++)printf("%d %d\n",lust,j);
        if(i<n)printf("%d %d\n",lust,i+1),lust=i+1;
        
    }
    return 0;
}
```