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> // ver2.6.3
//#include <complex.h> // FFT用
//#include <boost/math/tools/minima.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
//using boost::math::tools::brent_find_minima;//fmin(f,L,R,100)で最小値をmp(x,min)で取得
//using boost::multiprecision::cpp_int;//cpp_int (任意の長さのint)
#define fmin brent_find_minima
#define int long long
#define endl "\n"
#define ALL(v) (v).begin(),(v).end()
#define Vi vector<int>
#define VVi vector<Vi>
#define VVVi vector<VVi>
#define Vm vector<mint>
#define Vs vector<string>
#define Vd vector<double>
#define Pii pair<int,int>
#define Pdd pair<double,double>
#define VPii vector<Pii>
#define Tiii tuple<int,int,int>
#define VTiii vector<Tiii>
#define PQi priority_queue<int>
#define PQir priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define itos to_string
#define stoi stoll
#define FI first
#define SE second
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define cyes cout<<"yes"<<endl
#define cno cout<<"no"<<endl
#define sortr(v) sort(v,greater<>())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define dem(a,b) ((a+b-1)/(b))
#define Vin(a) rep(iI,0,a.size())cin>>a[iI]
#define INF 3000000000000000000 // 3.0*10^18(MAXの1/3くらい)
#define MAX LLONG_MAX
#define PI 3.141592653589793238462
//#define MOD 1000000007 // 10^9 + 7
int MOD = 998244353; // caseに応じて変える必要がある場合はこちら
const double EPS = 1e-10;

using namespace std;

struct mint{int x;mint(int x=0):x((x%MOD+MOD)%MOD){}
mint operator-()const{return mint(-x);}
mint& operator+=(const mint a){if((x+=a.x)>=MOD)x-=MOD;return*this;}
mint& operator-=(const mint a){if((x+=MOD-a.x)>=MOD)x-=MOD;return*this;}
mint& operator*=(const mint a){(x*=a.x)%=MOD;return*this;}
mint operator+(const mint a)const{return mint(*this)+=a;}
mint operator-(const mint a)const{return mint(*this)-=a;}
mint operator*(const mint a)const{return mint(*this)*=a;}
mint pow(int t)const{if(!t)return 1;mint a=pow(t>>1);a*=a;if(t&1)a*=*this;return a;}
mint inv()const{return pow(MOD-2);}
mint& operator/=(const mint a){return*this*=a.inv();}
mint operator/(const mint a)const{return mint(*this)/=a;}};
istream& operator>>(istream& is, const mint& a) {return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) {return os << a.x;}

/* デバッグ用 */
template <typename T> // vector出力
ostream& operator<<(ostream& os,const vector<T> &V){int N=V.size();
rep(i,0,N-1){os<<V[i]<<' ';}os<<V[N-1];return os;}
template <typename T> // vector<vector>出力
ostream& operator<<(ostream& os,const vector<vector<T>> &V){
int N=V.size();rep(i,0,N-1)cout<<V[i]<<endl;cout<<V[N-1];return os;}
template <typename T,typename S> // pair出力
ostream& operator<<(ostream& os, pair<T,S> const&P){os<<P.FI<<' '<<P.SE;return os;}
template <typename T> // set出力
ostream& operator<<(ostream& os,set<T> &S){auto it=S.begin();while(it!=S.end()){
os<<*it;if(next(it,1)!=S.end())os<<' ';it++;};return os;}
template <typename T> // deque出力
ostream& operator<<(ostream& os, deque<T> &q){for(auto it=q.begin();it<q.end();it++){
os<<*it;if(it+1!=q.end())os<<" ";}return os;}

/* 便利関数 */
int gcd(int a,int b){return b?gcd(b,a%b):a;} // 最大公約数gcd
int lcm(int a,int b){return a/gcd(a,b)*b;} // 最小公倍数lcm
int mypow(int x, int n, int m){ // 累乗x^n(mod m) O(log(n))
if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int scomb(int n, int r){if(r<0||r>n)return 0;if((n-r)<r)r=n-r; // nCr
int a=1;for(int i=n;i>n-r;--i){a=a*i;}for(int i=1;i<r+1;++i){a=a/i;}return a;}
Vi vis(Vi &v){Vi S(v.size()+1);rep(i,1,S.size())S[i]+=v[i-1]+S[i-1];return S;} // 累積和
int digit(int k,int i){string s = itos(k);return s[s.size()-i]-'0';} // i桁目の数字
void press(Vi &v){v.erase(unique(ALL(v)),v.end());} // 圧縮
Vi zip(Vi b){int Z=b.size(); // 座標圧縮
Pii p[Z+10];int a=b.size();Vi l(a);for(int i=0;i<a;i++)
p[i]=mp(b[i],i);sort(p,p+a);int w=0;for(int i=0;i<a;i++)
{if(i&&p[i].first!=p[i-1].first)w++;l[p[i].second]=w;}return l;}
Vi beki2V(){Vi a(63);int q = 1;rep(i,0,63){a[i]=q;q*=2;}return a;}
Vi beki2 = beki2V(); // 2^n配列
int modiv(int a,int b){return a*mypow(b,MOD-2,MOD)%MOD;} // a÷b(MOD環境)
int SMALLER(Vi &a,int x){return lower_bound(a.begin(),a.end(),x)-a.begin();}
int orSMALLER(Vi &a,int x){return upper_bound(a.begin(),a.end(),x)-a.begin();}
int BIGGER(Vi &a,int x){return a.size()-orSMALLER(a,x);}
int orBIGGER(Vi &a,int x){return a.size()-SMALLER(a,x);}
int COUNT(Vi &a,int x) {return upper_bound(ALL(a),x)-lower_bound(ALL(a),x);}
int maxind(Vi &a){return max_element(ALL(a))-a.begin();}
int minind(Vi &a){return min_element(ALL(a))-a.begin();}
Vi stpowV(){Vi a(300005);
a[0]=1;repeq(i,1,300004)a[i]=a[i-1]*i%MOD;return a;}
//Vi stpow = stpowV(); // 階乗配列(%MOD)

/****************************** START ******************************/

signed main() {
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(12);
  string s;
  cin >> s;
  if(s.back()=='1') {cout << -1 << endl;return 0;}
  if(s[0]=='0') {cout << -1 << endl;return 0;}
  int n = s.size();
  s.pop_back();
  string t = s;
  reverse(ALL(t));
  if(s!=t) {cout << -1 << endl;return 0;}
  
  int ct = 0;
  rep(i,0,s.size()-1){
    if(s[i]!=s[i+1]) ct++;
  }
  if(ct >= 3) {cout << -1 << endl;return 0;}
  
  int maxx = 0;
  rep(i,0,s.size()){
    if(s[i]=='1') maxx++;
    else break;
  }
  //cerr << maxx << endl;
  
  VPii ans;
  
  rep(i,0,n-1){
    if(i < maxx){
      ans.pb(mp(i+1,i+2));
    }
    else ans.pb(mp(1,i+2));
  }
  rep(i,0,n-1) cout << ans[i] << endl;
  return 0;
}