#include <bits/stdc++.h>
using namespace std;
#define N 200100

char s[N];
int n,last,cnt;
int u[N],v[N];

int main()
{
	scanf("%s",s+1); n=strlen(s+1);
	if (s[1]=='0' || s[n]=='1' || s[n-1]=='0') return puts("-1"),0;
	last=1;
	for (int i=2;i<n;i++) if (s[i]=='1')
	{
		if (s[n-i]=='0') return puts("-1"),0;
		for (int j=last;j<i;j++) {
			u[++cnt]=i;
			v[cnt]=j;
		}
		last=i;
	}
	for (int i=last;i<n;i++) u[++cnt]=n,v[cnt]=i;
	for (int i=1;i<n;i++) printf("%d %d\n",u[i],v[i]);
	return 0;
}
### Prompt

Please provide a cpp coded solution 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;
#define N 200100

char s[N];
int n,last,cnt;
int u[N],v[N];

int main()
{
	scanf("%s",s+1); n=strlen(s+1);
	if (s[1]=='0' || s[n]=='1' || s[n-1]=='0') return puts("-1"),0;
	last=1;
	for (int i=2;i<n;i++) if (s[i]=='1')
	{
		if (s[n-i]=='0') return puts("-1"),0;
		for (int j=last;j<i;j++) {
			u[++cnt]=i;
			v[cnt]=j;
		}
		last=i;
	}
	for (int i=last;i<n;i++) u[++cnt]=n,v[cnt]=i;
	for (int i=1;i<n;i++) printf("%d %d\n",u[i],v[i]);
	return 0;
}
```