Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

1111

Sample Output 1

-1

It is impossible to have a connected component of size n after removing one edge from a tree with n vertices.

Sample Input 2

1110

Sample Output 2

1 2
2 3
3 4

If Edge 1 or Edge 3 is removed, we will have a connected component of size 1 and another of size 3. If Edge 2 is removed, we will have two connected components, each of size 2.

Sample Input 3

1010

Sample Output 3

1 2
1 3
1 4

Removing any edge will result in a connected component of size 1 and another of size 3.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
char s[100010];
int main()
{
	int i,qz,nn,rt,n;
	scanf("%s",s+1);n=strlen(s+1);
	if(s[n]=='1'||s[1]=='0'){printf("-1");return 0;}
	for(i=1;i<n;i++)
	{
		if(s[i]!=s[n-i]){printf("-1");return 0;}
	}
	if(n%2==0)nn=n/2;
	else nn=n/2+1;
	qz=0;for(i=1;i<=nn;i++)if(s[i]=='1')qz++;else break;
	for(i=2;i<=nn;i++)if(s[i]=='1')printf("%d %d\n",i-1,i);else break;
	rt=qz;
	for(i=qz+1;i<=nn;i++)
	{
		if(s[i]=='0')
		{
			if(s[i-1]=='1')
			{
				printf("%d %d\n",rt,i);
				rt=i;
			}
			else
			{
				printf("%d %d\n",rt,i);
			}
		}
		else
		{
			if(s[i-1]=='0')
			{
				printf("%d %d\n",rt,i);
			}
			else
			{
				printf("%d %d\n",rt,i);
				rt=i;
			}
		}
	}
	if(s[nn]=='0')
	{
		for(i=nn+1;i<=n;i++)printf("%d %d\n",rt,i);
	}
	else
	{
		printf("%d %d\n",rt,nn+1);
		for(i=nn+2;i<=n;i++)printf("%d %d\n",nn+1,i);
	}
	return 0;
}