#include<bits/stdc++.h>
#define ll long long
#define rg register
using namespace std;
const int MAXN=1e5+10;
char S[MAXN];
int ls,Root,Ds;
int main()
{   scanf("%s",S+1),ls=strlen(S+1);
	if(S[ls]=='1'||S[1]=='0') {   puts("-1");   return 0;   }
	for(rg int i=1;i<ls;i++)
		if(S[i]!=S[ls-i]){   puts("-1");   return 0;   }
	puts("1 2"),Root=2;
	for(rg int i=3;i<=ls;i++)
	{   printf("%d %d\n",Root,i);
		if(S[i-1]=='1') Root=i;
	}
}

### Prompt

Please create a solution in CPP 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 ll long long
#define rg register
using namespace std;
const int MAXN=1e5+10;
char S[MAXN];
int ls,Root,Ds;
int main()
{   scanf("%s",S+1),ls=strlen(S+1);
	if(S[ls]=='1'||S[1]=='0') {   puts("-1");   return 0;   }
	for(rg int i=1;i<ls;i++)
		if(S[i]!=S[ls-i]){   puts("-1");   return 0;   }
	puts("1 2"),Root=2;
	for(rg int i=3;i<=ls;i++)
	{   printf("%d %d\n",Root,i);
		if(S[i-1]=='1') Root=i;
	}
}

```