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: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
#define reg register
const int MN=1e5+5;
char s[MN];
int n,rt;bool a[MN];
int main()
{
	scanf("%s",s+1);n=strlen(s+1);
	reg int i;
	for(i=1;i<=n;++i) a[i]=(s[i]=='1');
	if(!a[1]||a[n]) return 0*puts("-1");
	for(i=1;i<=n;++i) if(a[i]!=a[n-i]) return 0*puts("-1");
	for(rt=1,i=2;i<=(n+1)/2;++i)
	{
		if(a[i]&&a[i-1]) printf("%d %d\n",i,rt),rt=i;
		if(a[i]&&!a[i-1])
		{
			printf("%d %d\n",i,rt);
			for(int j=rt+1;j<i;++j) printf("%d %d\n",i,j);
		}
	}
	printf("%d %d\n",rt+1,rt);++rt;
	for(i=rt+1;i<=n;++i) printf("%d %d\n",rt,i);
	return 0;
}