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;


#define li          long long int
#define rep(i,to)   for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)    for(li i=(li)(start);i<((li)(to));i++)
#define pb          push_back
#define sz(v)       ((li)(v).size())
#define bgn(v)      ((v).begin())
#define eend(v)     ((v).end())
#define allof(v)    (v).begin(), (v).end()
#define dodp(v,n)       memset(v,(li)n,sizeof(v))
#define bit(n)      (1ll<<(li)(n))
#define mp(a,b)     make_pair(a,b)
#define rin rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)     idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)       idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)        idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)   for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)      idp p2((x).F, (x).S)
#define dump(x,n)   idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)  idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)    idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)   idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)   rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)   rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;

inline bool ok(string& s) {
    li n = sz(s);
    if (s[0] == '0') {
        return false;
    }
    if (s[n - 1] == '1') {
        return false;
    }
    rep(i, (n - 1) / 2) {
        if (s[i] != s[n - 2 - i]) {
            return false;
        }
    }
    return true;
}

int main() {
    string s;
    cin >> s;
    li n = sz(s);
    if (!ok(s)) {
        puts("-1");
        return 0;
    }
    if (n == 2) {
        puts("1 2");
        return 0;
    }

    li largest_tree = 0;
    rep(i, n / 2) {
        if (s[i] == '1') {
            largest_tree = i;
        }
    }
    if (largest_tree == 0) {
        rep(i, n - 1) {
            cout << 1 << " " << i + 2 << endl;
        }
        return 0;
    }
    li head = 1;
    rep(i, largest_tree) {
        cout << head << " " << i + 2 << endl;
        cout << n - head + 1 << " " << n - i - 1 << endl;
        if (s[i] == '1') {
            head = i + 2;
        }
    }
    li ahead = n + 1 - head;
    cout << head << " " << ahead << endl;
    repp(i, largest_tree + 1, n - largest_tree - 1) {
        cout << head + 1 << " " << i + 1 << endl;
    }

    return 0;
}