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<string>
#include<iostream>
#include<vector>
using namespace std;
vector<pair<int, int>> vt;
int check[100010];
int main()
{
	string a;
	cin >> a;
	int t = a.length();
	if (a[0] == '0' || a[t - 1] == '1' || a[t - 2] == '0')
	{
		puts("-1"); return 0;
	}
	if (t == 2)
	{
		printf("1 2");
		return 0;
	}
	int update = 0;
	for (int i = 0; i < t-1; i++)
	{
		if (a[i] != a[t - 2 - i])
		{
			puts("-1"); return 0;
		}
		if (a[i] == '0')update = 1;
		if (update&&a[i] == '1'&&!check[i])
		{
			puts("-1"); return 0;
		}
		if (update)
		{
			vt.push_back({ 1,i + 2 });
		}
		if (a[i] == '1')
			vt.push_back({ i+1,i + 2 });
		check[i] = 1;
		check[t - 2 - i] = 1;
	}
	for (int i = 0; i < vt.size(); i++)
		printf("%d %d\n", vt[i].first, vt[i].second);
}