#include<bits/stdc++.h>
using namespace std;
#define maxn 100010
#define LL long long
#define inf 0x3f3f3f3f

int n,m,k;
int a[maxn];
string s;
int root;

int main(){
//	freopen("test.txt","r",stdin);
	cin>>s;
	n=s.length();
	for(int i=1;i<=n;i++) a[i]=(s[i-1]=='1');
	for(int i=1;i<=n;i++)
		if(a[i]!=a[n-i]){
			puts("-1");
			return 0;
		}
	if(a[1]==0||a[n-1]==0||a[n]==1){
		puts("-1");
		return 0;
	}
	root=1;
	for(int i=1;i<n;i++){
		printf("%d %d\n",root,i+1);
		if(a[i]==1){
			root=i+1;
		}else{
			
		}
	}
	
    return 0;
}
### Prompt

Your task is to create 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;
#define maxn 100010
#define LL long long
#define inf 0x3f3f3f3f

int n,m,k;
int a[maxn];
string s;
int root;

int main(){
//	freopen("test.txt","r",stdin);
	cin>>s;
	n=s.length();
	for(int i=1;i<=n;i++) a[i]=(s[i-1]=='1');
	for(int i=1;i<=n;i++)
		if(a[i]!=a[n-i]){
			puts("-1");
			return 0;
		}
	if(a[1]==0||a[n-1]==0||a[n]==1){
		puts("-1");
		return 0;
	}
	root=1;
	for(int i=1;i<n;i++){
		printf("%d %d\n",root,i+1);
		if(a[i]==1){
			root=i+1;
		}else{
			
		}
	}
	
    return 0;
}
```