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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}

int main() {
    string s;
    cin >> s;
    int n = s.size();
    if(s[n-1]=='1'){
        cout << "-1" << endl;
        return 0;
    }
    if(s[0]=='0'){
        cout << "-1" << endl;
        return 0;
    }
    rep(i,n/2){
        if(s[i]!=s[n-2-i]){
            cout << "-1" << endl;
            return 0;
        }
    }

    int now = 1;
    srep(i,1,n){
        cout << now << ' ' << i+1 << endl;
        if(n%2==0){
            if(n/2<=i){
                if(s[n-1-i]=='1'){
                    now = i+1;
                }
            }
        }else{
            if((n+1)/2<=i){
                if(s[n-1-i]=='1'){
                    now = i+1;
                }
            }
        }
    }

    return 0;
}