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)

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define                           int                         long long
#define                           fi                          first
#define                           si                          second
#define                           mp                          make_pair
#define                           pb                          push_back
#define                           pi                          pair<int,int>
#define                           f(i,l,r)                    for(int i=l;i<=r;i++)
#define                           rf(i,r,l)                   for(int i=r;i>=l;i--)
#define                           done(i)                     cout<<"done = "<<i<<endl;
#define                           fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);




using namespace std;

const int inf=1e18;
const int mod=1e9+7;
const int M=100009;
inline int bigmod(int B,int P){int R=1;while(P>0){if(P&1){R=(R*B)%mod;}P>>=1;B=(B*B)%mod;}return R;}


 main()

{
    fast
    string s;
    cin>>s;
    int sz=s.size();
    if(s[sz-1]=='1' || s[0]=='0')
    {
        cout<<"-1"<<endl;
        return 0;
    }
    int mid=sz/2;
    f(i,0,mid-1)
    {
        int lft=i;
        int rgt=sz-(i+2);
        if(s[lft]!=s[rgt])
        {
            cout<<"-1"<<endl;
            return 0;
        }
    }
    int cur=1;
    int nxt=2;
    int pos=0;
    vector<pi>ses;
    while(pos<mid)
    {
        if(s[pos]=='1')
        {
            ses.pb(mp(cur,nxt));
            cur=nxt;nxt++;
        }
        else
        {
           ses.pb(mp(cur,nxt));
           nxt++;
        }
        pos++;
    }
    f(i,nxt,sz)
    {
        ses.pb(mp(cur,i));
    }
    for(auto x:ses)
    {
        cout<<x.fi<<" "<<x.si<<endl;
    }
    return 0;

}