#include<bits/stdc++.h>
using namespace std;
vector<int> v;
int tot=1;
void build(int u,int pos)
{
	if(!pos)return;
	printf("%d %d\n",u,++tot);
	build(tot,pos-1);
	for(int i=v[pos-1]+1;i<v[pos];i++)printf("%d %d\n",u,++tot);
}
string s;
int n;
int main()
{
	cin>>s;
	n=int(s.size());s='0'+s;
	if(s[1]=='0')return puts("-1"),0;
	for(int i=0;i<=n;i++)
		if(s[i]!=s[n-i])return puts("-1"),0;
		else if(s[i]=='1')v.push_back(i);
	v.push_back(n);
	build(1,int(v.size()-1));
	assert(tot==n);
}

### Prompt

Develop a solution in CPP 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;
vector<int> v;
int tot=1;
void build(int u,int pos)
{
	if(!pos)return;
	printf("%d %d\n",u,++tot);
	build(tot,pos-1);
	for(int i=v[pos-1]+1;i<v[pos];i++)printf("%d %d\n",u,++tot);
}
string s;
int n;
int main()
{
	cin>>s;
	n=int(s.size());s='0'+s;
	if(s[1]=='0')return puts("-1"),0;
	for(int i=0;i<=n;i++)
		if(s[i]!=s[n-i])return puts("-1"),0;
		else if(s[i]=='1')v.push_back(i);
	v.push_back(n);
	build(1,int(v.size()-1));
	assert(tot==n);
}

```