#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll N=2e5+5;
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    string s;cin>>s;
    ll n = s.size();
    s="0" + s;
    if(s[n]=='1' || s[1]=='0')return cout<<"-1" , 0;
    for(ll i=1;i<=n-1;i++)if(s[i]!=s[n-i])return cout<<"-1" , 0;
    for(ll i=1,u=1,prv=0;i<=n-1;i++){
        if(s[i]=='0')continue;
        ll v=u+i-prv;
        for(ll j=u+1;j<=v;j++)cout<<u<<" "<<j<<"\n";
        u=v,prv=i;
    }
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
typedef long long ll;
using namespace std;
const ll N=2e5+5;
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    string s;cin>>s;
    ll n = s.size();
    s="0" + s;
    if(s[n]=='1' || s[1]=='0')return cout<<"-1" , 0;
    for(ll i=1;i<=n-1;i++)if(s[i]!=s[n-i])return cout<<"-1" , 0;
    for(ll i=1,u=1,prv=0;i<=n-1;i++){
        if(s[i]=='0')continue;
        ll v=u+i-prv;
        for(ll j=u+1;j<=v;j++)cout<<u<<" "<<j<<"\n";
        u=v,prv=i;
    }
}
```