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 <iostream>
#include <fstream>
#include <iomanip>
#include <climits>
#include <limits>
#include <algorithm>
#include <array>
#include <vector>
#include <deque>
#include <queue>
#include <list>
#include <stack>
#include <string>
#include <functional>
#include <numeric>
#include <map>
#include <set>
#include <cstdlib>
#include <bitset>
#include <unordered_map>
#include <random>
#include <cmath>
#include <complex>
// #include "utiltime.hpp"

using namespace std;

typedef long long int ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> P;
typedef pair<ll, ll> Pll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef complex<double> cdouble;

const int INFL = (int)1e9;
const ll INFLL = (ll)1e18;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
#define Loop(i, n) for(int i = 0; i < (int)n; i++)
#define Loopll(i, n) for(ll i = 0; i < (ll)n; i++)
#define Loop1(i, n) for(int i = 1; i <= (int)n; i++)
#define Loopll1(i, n) for(ll i = 1; i <= (ll)n; i++)
#define Loopr(i, n) for(int i = (int)n - 1; i >= 0; i--)
#define Looprll(i, n) for(ll i = (ll)n - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)n; i >= 1; i--)
#define Looprll1(i, n) for(ll i = (ll)n; i >= 1; i--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define printv(vector) Loop(i, vector.size()) { cout << vector[i] << " "; } cout << endl;
#define printmx(matrix) Loop(i, matrix.size()) { Loop(j, matrix[i].size()) { cout << matrix[i][j] << " "; } cout << endl; }
#define quickio() ios::sync_with_stdio(false); cin.tie(0);
#define readfile(filename) ifstream in(filename); cin.rdbuf(in.rdbuf());
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
bool nearlyeq(double x, double y) { return abs(x - y) < 1e6; }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(double x) { ll m = (ll)sqrt(x); return m + (m * m <= (ll)(x) ? 0 : -1); }
ll ceilsqrt(double x) { ll m = (ll)sqrt(x); return m + ((ll)x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }

/*******************************************************/

void failed() {
	cout << -1 << endl;
	exit(0);
}

int main() {
	string s; cin >> s;
	int n = s.length();
	if (s[0] == '0' || s[n - 1] == '1') failed();
	s.pop_back();
	string t = s;
	reverse(t.begin(), t.end());
	if (s != t) failed();
	vector<P> ans;
	if (n == 2) ans = { {1,2} };
	else if (n == 3) ans = { {1,2},{2,3} };
	else {
		vi v = { 1 };
		Loop1(i, n / 2 - 1) {
			if (s[i] == '0') {
				v.push_back(i + 1);
			}
			else {
				Loop(j, v.size()) {
					ans.push_back({ v[j], i + 1 });
				}
				v = { i + 1 };
			}
		}
		Loop(j, v.size()) {
			ans.push_back({ v[j], n / 2 + 1 });
		}
		for (int i = n / 2 + 1; i < n; ++i) {
			ans.push_back({ i + 1, n / 2 + 1 });
		}
	}
	Loop(i, ans.size()) {
		cout << ans[i].first << " " << ans[i].second << endl;
	}
}