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 <cmath>
#include <iostream>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <algorithm>
#include <functional>
#include <utility>
#include <string>
#define typeof(x) __typeof__(x)
#define int long long int
#define mod(x) ((x % MOD) + MOD) % MOD
#define ceil(x,m) ((x)+(m)-1)/(m)
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);--i)

#define ALL(c) (c).begin(),(c).end()
#define RALL(c) (c).rbegin(),(c).rend()
#define SZ(c) (int)((c).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define SORT(c) sort(ALL(c))
#define RSORT(c) sort(RALL(c))
#define LB(c,x) (int)(lower_bound(ALL(c),x)-(c).begin())
#define UB(c,x) (int)(upper_bound(ALL(c),x)-(c).begin())
#define COUNT(c,x) UB(c,x)-LB(c,x)
#define UNIQUE(c) SORT(c); (c).erase(unique(ALL(c)),(c).end());
#define COPY(c1,c2) copy(ALL(c1),(c2).begin())
#define EXIST(s,e) (s).find(e)!=(s).end()
#define PB push_back
#define MP make_pair
#define vec vector

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl;

using namespace std;

typedef pair<int,int> P;
struct edge { int to, cost; };

const int INF = 1e18;
const int MOD = 1e9+7;

template<typename T> ostream& operator << (ostream& s, const vector<T>& v) {
  int len = v.size();
  s << "[";
  for (int i = 0; i < len; i++) { s << v[i]; if (i < len - 1) s << " "; }
  s << "]";
  return s;
}

signed main()
{
  string s; cin >> s;
  int n = SZ(s);

  if (s[0] == '0' || s[n-1] == '1') {
    cout << -1 << endl;
    return 0;
  }
  rep(i, 0, n-1) {
    if (s[i] != s[n-2-i]) {
      cout << -1 << endl;
      return 0;
    }
  }

  vec< pair<int,int> > E;
  int cur = 0;
  rep(i, 0, n-1) {
    E.PB(MP(cur, i+1));
    if (s[i] == '1') {
      cur = i+1;
    }
  }

  rep(i, 0, SZ(E)) {
    cout << E[i].first+1 << " " << E[i].second+1 << endl;
  }

  return 0;
}