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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

#define forn(i, s, f) for (int i = s; i < f; i++)
#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair <int, int>
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    string s;
    cin >> s;
    if (s.back() == '1' || s[0] == '0')
        return cout << -1, 0;
    s.pop_back();
    int n = sz(s);
    forn (i, 0, n)
        if (s[i] != s[n - i - 1])
            return cout << -1, 0;
    vector <int> a;
    forn (i, 0, n)
        if (s[i] == '1')
            a.pb(i);
    forn (i, 0, sz(a))
        cout << i + 1 << " " << i + 2 << "\n";
    int cnt = sz(a) + 2;
    forn (i, 1, sz(a)) {
        forn (j, 0, a[i] - a[i - 1] - 1) {
            cout << cnt++ << " " << i + 1 << "\n";
        }
    }
    return 0;
}