#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;

int main(){
    string s;cin>>s;
    bool ok=(s[s.size()-1]=='0')&(s[0]=='1');
    lol(i,s.size()-1)ok&=(s[i]==s[s.size()-i-2]);
    if(!ok)return cout<<-1<<endl,0;
    ll bef=mod;
    lol(i,s.size()){
	if(s[i]=='0')continue;
	for(ll j=bef;j<i;j++)cout<<j+1<<" "<<i+1<<endl;
	bef=i;
    }
    return cout<<s.size()-1<<" "<<s.size()<<endl,0;
}

### Prompt

In CPP, your task is to solve 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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;

int main(){
    string s;cin>>s;
    bool ok=(s[s.size()-1]=='0')&(s[0]=='1');
    lol(i,s.size()-1)ok&=(s[i]==s[s.size()-i-2]);
    if(!ok)return cout<<-1<<endl,0;
    ll bef=mod;
    lol(i,s.size()){
	if(s[i]=='0')continue;
	for(ll j=bef;j<i;j++)cout<<j+1<<" "<<i+1<<endl;
	bef=i;
    }
    return cout<<s.size()-1<<" "<<s.size()<<endl,0;
}

```