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>
#include<limits.h>
#include<stdio.h>
#include<cstring>
#include<string>
using namespace  std;
typedef long long ll;
typedef long double ld;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define sz(a) a.size()
#define fi first
#define se second
#define rsz(a,n) a.resize(n)
#define test(t) ll t;cin>>t;while(t--)
#define forn(i,e) for(int i = 0; i < e; i++)
#define forsn(i,s,e) for(int i = s; i < e; i++)
#define rforn(i,s) for(int i = s; i >= 0; i--)
#define rforsn(i,s,e) for(int i = s; i >= e; i--)
#define fillv(a,k) memset(a,k,sizeof(a))
#define leadzero(a) __builtin_clz(a) //count leading zeros
#define trailzero(a) __builtin_ctz(a) //count trailing zeros
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln cout<<"\n"
#define sp cout<<" "
#define spaceprint(a,i,s,n) {forsn(i,s,n) { cout<<a[i]; sp;}}
#define lineprint(a,i,s,n) {forsn(i,s,n) {cout<<a[i]; ln;}}
#define maxe(a) *max_element(a.begin(),a.end())
#define maxi(a) max_element(a.begin(),a.end())-a.begin()
#define mine(a) *min_element(a.begin(),a.end())
#define mini(a) min_element(a.begin(),a.end())-a.begin()
#define all(c) c.begin(),c.end()
#define trav(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define present(container, element) (container.find(element) != container.end())
#define cpresent(container, element)  (find(all(container),element) != container.end())// check the presence of element
//copy(from_begin, from_end, to_begin); copy function
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<string> vs;
typedef vector<int> v32;
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<p64> vp64;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;
typedef map<ll,ll> m64;
const int LIM = 1e5+5, MOD = 998244353;
#define sumv(v) accumulate(all(v),ll(0))
#define productv(v) accumulate(all(v), ll(1), multiplies< ll >())
ll gcd(ll a, ll b) { if(b == 0) return a; return gcd(b, a % b); }
ll fastpowMOD(ll a, ll p,ll MOD){ if(p==0) return 1; ll z = fastpowMOD(a,p/2,MOD); z = (z*z)%MOD; if(p%2) z = (z*a)%MOD; return z; }
bool seive[2005];
void SieveOfEratosthenes(ll n)
{ memset(seive, true, sizeof(seive)); for (ll p=2; p*p<=n; p++) if (seive[p] == true) for (ll i=p*p; i<=n; i += p) seive[i] = false; }
ll fastpow(ll a, ll p){ if(p==0) return 1; ll z = fastpow(a,p/2); z = (z*z); if(p%2) z = (z*a); return z; }
ll a[3*LIM],b[3*LIM],m,n;
int main()
{
    fast;
    ll n,m;
    char s[100005];
    cin>>s+1;
    n=strlen(s+1);
    //cout<<n;
    v64 v;
    ll i;
    ll flag=0;
    if(s[1]=='0')
        flag=1;
    if(s[n]=='1')
        flag=1;
    for(i=1;i<n;i++)
    {
        if(s[i]=='1')
            v.pb(i);
        if(s[i]!=s[n-i])
        {
            flag=1;
            break;
        }
    }

    if(flag==1)
    {
        cout<<-1;
        return 0;
    }
    if(n==2)
    {
        cout<<1<<" "<<2;
    }
    vp64 edge;
    ll prev=1;
    for(auto it :v)
    {
        if(it==1)
            continue;
        edge.pb(mp(prev,it));
        for(i=prev+1;i<it;i++)
        {
            edge.pb(mp(i,it));
        }
        prev=it;
    }
    edge.pb(mp(n,n-1));
    for(auto it:edge)
    {
        cout<<min(it.fi,it.se)<<" "<<max(it.se,it.fi);
        ln;
    }


}