#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
string s;
ll n;
queue<ll> q;
int main(){
	cin >> s;
	n=s.size();
	if(s[0]!='1' or s[n-1]!='0')
		return puts("-1"),0;
	for(int i=0;i<n and n-1-i>0;i++)
		if(s[i]!=s[n-2-i])
			return puts("-1"),0;
	for(int i=0;i<n;i++){
		if(s[i]=='1' or i==n-1)
			while(!q.empty()){
				printf("%lld %d\n",q.front()+1,i+1);
				q.pop();
			}
		q.push(i);
	}
}
### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
string s;
ll n;
queue<ll> q;
int main(){
	cin >> s;
	n=s.size();
	if(s[0]!='1' or s[n-1]!='0')
		return puts("-1"),0;
	for(int i=0;i<n and n-1-i>0;i++)
		if(s[i]!=s[n-2-i])
			return puts("-1"),0;
	for(int i=0;i<n;i++){
		if(s[i]=='1' or i==n-1)
			while(!q.empty()){
				printf("%lld %d\n",q.front()+1,i+1);
				q.pop();
			}
		q.push(i);
	}
}
```