#include<bits/stdc++.h>
using namespace std;
const int N=100005;
char s[N];
int a[N];
void addedge(int u,int v){
	printf("%d %d\n",u,v);
}
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	for (int i=1;i<=n;i++)
	a[i]=s[i]-'0';
	if (a[1]==0||a[n]==1){printf("-1\n"); return 0;}
	for (int i=1;i<=n;i++)
	if ((a[i]^a[n-i])==1){printf("-1\n"); return 0;}
	int now=1;
	for (int i=2;i<=n;i++){
		addedge(now,i);
		if (a[i-1]==1) now=i;
	}
	return 0;
}
### Prompt

Create a solution in Cpp for 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;
const int N=100005;
char s[N];
int a[N];
void addedge(int u,int v){
	printf("%d %d\n",u,v);
}
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	for (int i=1;i<=n;i++)
	a[i]=s[i]-'0';
	if (a[1]==0||a[n]==1){printf("-1\n"); return 0;}
	for (int i=1;i<=n;i++)
	if ((a[i]^a[n-i])==1){printf("-1\n"); return 0;}
	int now=1;
	for (int i=2;i<=n;i++){
		addedge(now,i);
		if (a[i-1]==1) now=i;
	}
	return 0;
}
```