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<iostream>
#include<algorithm>
#include<climits>
#include<vector>
#include<utility>//pair
#include<cmath>//abs,sqrt
#include<cstring>
#include<map>
#include<string>
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define prin(arg) cout<<arg<<"\n"
#define prin2(arg1,arg2) cout<<arg1<<" "<<arg2<<"\n"
#define fill(arg,n) memset(arg,n,sizeof(arg))
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
const int INF=1e+9;
const ll INFLL=1e+17;
string s;
void solve(){
  int n=s.size();
  if(s[0]=='0'||s[n-1]=='1'){
    prin(-1);
    return;
  }
  rep(i,n/2){
    if(s[i]!=s[n-2-i]){
      prin(-1);
      return;
    }
  }
  int cnt=1;
  rep1(i,n/2-1){
    if(s[i]=='1'){
      rep1(j,cnt){
	prin2(i+1,i+1-j);
      }
      cnt=1;
    }
    else{
      cnt++;
    }
  }
  if(s[n/2-1]=='0'){
    rep1(j,cnt-1){
      prin2(n/2,n/2-j);
    }
    rep(i,n/2){
      prin2(n/2,n/2+1+i);
    }
  }
  else{
    prin2(n/2,n/2+1);
    rep(i,n/2-1){
      prin2(n/2+1,n/2+2+i);
    }
  }
}
int main(){
  cin>>s;
  solve();
  return 0;
}