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 <cstdio>
#include <cstring>
#include <cmath>
#include <bitset>
#include <iostream>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn=100100;
int n;
char sz[maxn];
int main(){
    scanf("%s",sz+1);
    n=strlen(sz+1);
    if(sz[n]=='1'||sz[1]=='0'||sz[n-1]=='0'){
        printf("-1\n");
        return 0;
    }
    for(int i=2;i<=n-1;i++){
        if(sz[i]!=sz[n-i]){
            printf("-1\n");
            return 0;
        }
    }
    int root=2,num=1,now=2;
    printf("2 1\n");
    while(now<=n-now){
        if(sz[now]=='1'){
            printf("%d %d\n",++now,root);
            root=now;
        }else{
            printf("%d %d\n",++now,root);
        }
        num++;
    }
    for(int i=num+1;i<n;i++){
        printf("%d %d\n",root,++now);
    }
    return 0;
}