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<bits/stdc++.h>
using namespace std;

const int N = 1e5;

string s;
int par[N+5];
vector<int> adj[N+5];
bool vis[N+5];
int le,ri;
int cnt = 0;

void dfs(int node, int id1, int id2)
{
//   cout << "node = " << node << " >> " << id1 << " " << id2 << endl;
   for(int i=0;i<adj[node].size();i++)
   {
      int now = adj[node][i];
      if(vis[now]) continue;
      vis[now] = 1;
      cnt++;
      par[le+cnt] = id1;
      par[ri+cnt] = id2;
      dfs(now,le+cnt,ri+cnt);
   }
   return;
}

int main()
{
   ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
   cin >> s;
   int pan = s.size()-1;
   bool bisa=1;
   if(s[pan]=='1' || s[0] =='0') bisa=0;
   pan--;
   for(int i=0;i<pan/2;i++)
   {
      if(s[i]!=s[pan-i]) bisa=0;
   }
   if(!bisa) printf("-1\n");
   else
   {
      int cur=1;
      for(int i=0;i<=(pan+1)/2;i++)
      {
         if(s[i] == '1')
         {
            adj[cur].push_back(i+2);
            adj[i+2].push_back(cur);
            cur=i+2;
         }
         else
         {
            adj[i+2].push_back(cur);
            adj[cur].push_back(i+2);
         }
//         cout << "current root = " << cur << endl;
      }
      le = 1;
      pan+=2;
      if(pan%2==0)
      {
         ri = ((pan)/2)+1;
         par[ri] = le;
      }
      else
      {
         ri = (pan+1)/2;
      }
      vis[cur] = 1;
//      cout << ">> " << le << " " << ri << endl;
      dfs(cur,le,(pan%2==0 && s[pan/2-1]=='1')?ri:1);
      for(int i=2;i<=pan;i++)
      {
         printf("%d %d\n",i,par[i]);
      }
   }
 	return 0;
}