//Copyright(c)2018 Mstdream
#include<bits/stdc++.h>
using namespace std;
const int N=100010;
char s[N];
void add(int x,int y){printf("%d %d\n",x,y);}
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	for(int i=1;i<=n;i++)s[i]-='0';
	bool flag=1;
	if(s[1]==0||s[n-1]==0||s[n]==1)flag=0;
	for(int i=1;i<=n;i++)if(s[i]^s[n-i])flag=0;
	if(!flag)puts("-1"),exit(0);
	int p=1,pre=0;
	for(int i=2;i<=n;i++){
		if(s[i]){
			if(pre)add(pre,i);
			for(int j=pre+1;j<i;j++)add(i,j);
			pre=i;
		}
	}
	for(int i=pre+1;i<=n;i++)add(pre,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
//Copyright(c)2018 Mstdream
#include<bits/stdc++.h>
using namespace std;
const int N=100010;
char s[N];
void add(int x,int y){printf("%d %d\n",x,y);}
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	for(int i=1;i<=n;i++)s[i]-='0';
	bool flag=1;
	if(s[1]==0||s[n-1]==0||s[n]==1)flag=0;
	for(int i=1;i<=n;i++)if(s[i]^s[n-i])flag=0;
	if(!flag)puts("-1"),exit(0);
	int p=1,pre=0;
	for(int i=2;i<=n;i++){
		if(s[i]){
			if(pre)add(pre,i);
			for(int j=pre+1;j<i;j++)add(i,j);
			pre=i;
		}
	}
	for(int i=pre+1;i<=n;i++)add(pre,i);
}

```