#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define pii pair<int,int>
using namespace std;
const int N=1e5+5;
int n,cnt; pii e[N]; char s[N];
int main(){
	scanf("%s",s+1); n=strlen(s+1);
	if (s[n]=='1'||s[1]=='0') return puts("-1"),0;
	rep (i,1,n/2) if (s[i]^s[n-i]) return puts("-1"),0;
	int x=1,y=1;
	rep (i,1,n-1){ e[++cnt]=pii(x,++y); if (s[i]=='1') x=y; }
	rep (i,1,n-1) printf("%d %d\n",e[i].first,e[i].second);
	return 0;
}
### 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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define pii pair<int,int>
using namespace std;
const int N=1e5+5;
int n,cnt; pii e[N]; char s[N];
int main(){
	scanf("%s",s+1); n=strlen(s+1);
	if (s[n]=='1'||s[1]=='0') return puts("-1"),0;
	rep (i,1,n/2) if (s[i]^s[n-i]) return puts("-1"),0;
	int x=1,y=1;
	rep (i,1,n-1){ e[++cnt]=pii(x,++y); if (s[i]=='1') x=y; }
	rep (i,1,n-1) printf("%d %d\n",e[i].first,e[i].second);
	return 0;
}
```