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 <stdio.h>
#include <string.h>
#include <vector>
#include <utility>

using namespace std;

typedef pair<int, int> ip;

char s[100002];
int n;

int main() {
	scanf("%s", &s[1]);
	n = strlen(&s[1]);
	if (s[1] != '1' || s[n - 1] != '1' || s[n] != '0') {
		puts("-1");
		return 0;
	}
	for (int i = 1; i <= n - 1; i++) {
		if (s[i] != s[n - i]) {
			puts("-1");
			return 0;
		}
	}
	vector<ip> res;
	int root = 1, cur = 2;
	for (int i = 2; i <= n; i++) {
		res.push_back(ip(cur++, root));
		if (s[i] == '1') root = cur - 1;
	}
	for (int i = 0; i < res.size(); i++)
		printf("%d %d\n", res[i].first, res[i].second);
	return 0;
}