#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	string S; cin >> S;
	int N = S.size();
	if(S[0] == '0' || S[N-1] == '1') return cout << -1 << endl, 0;
	repp(i,0,N-1) if(S[i] != S[N-2-i]) return cout << -1 << endl, 0;
	int z = 1;
	repp(i,1,N) if(S[i] == '1'){
		while(z <= i){
			cout << z << ' ' << i+1 << endl;
			++z;
		}
	}
	repp(i,z+1,N+1) cout << z << ' ' << i << endl;
	return 0;
}

### Prompt

Develop a solution in cpp 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;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	string S; cin >> S;
	int N = S.size();
	if(S[0] == '0' || S[N-1] == '1') return cout << -1 << endl, 0;
	repp(i,0,N-1) if(S[i] != S[N-2-i]) return cout << -1 << endl, 0;
	int z = 1;
	repp(i,1,N) if(S[i] == '1'){
		while(z <= i){
			cout << z << ' ' << i+1 << endl;
			++z;
		}
	}
	repp(i,z+1,N+1) cout << z << ' ' << i << endl;
	return 0;
}

```