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 <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>
#include <string>
#include <cstring>

#define sz(v) ((int)((v).size()))
#define all(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int INF = 0x3c3c3c3c;
const ll LINF = 1ll*INF*INF*2;

const int MAX_N = 100001;
int n;
string s;
bool input[MAX_N];
vector<int> v;

int main(void) {
	cin >> s;
	n = s.size();
	for(int i=0; i<n; i++) input[i+1] = s[i]-'0';
	for(int i=1; i<n; i++) if(input[n-i] != input[i]) {printf("-1"); return 0;}
	if(input[n]) {printf("-1"); return 0;}
	if(!input[1]) {printf("-1"); return 0;}
	//10이면 5까지 , 9면 4까지

	for(int i=1; i<=n/2; i++) if(input[i]) v.push_back(i);
	int szv = sz(v);

	int cur=1;
	for(int i=1; i<szv; i++) {
		int temp = cur+1;

		while(temp<=v[i]) {
			printf("%d %d\n", cur, temp);
			temp++;
		}
		printf("%d %d\n", cur, temp);
		cur = temp;
	}
	for(int temp = cur+1; temp<=n; temp++) printf("%d %d\n", cur, temp);
}