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 <cassert>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
using namespace std;
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
typedef long long ll;


char buf[100000+10];
int n;

bool isOK(void) {
  if(buf[n-1] == '1') return false; // 0とnには分けられない
  if(buf[0] == '0' || buf[n-2] == '0') return false; // 葉はある
  REP(i, n-1) {
    if(buf[i] == '1' && buf[n-2-i] == '0') {
      return false;
    }
  }
  return true;
}

int main(void) {
  scanf("%s", buf);
  n = strlen(buf);
  if(!isOK()) {
    puts("-1");
    return 0;
  }
  printf("1 2\n");
  int u = 2; // 今いるところ
  int v = 3; // 次に繋ぐ頂点
  for(int i = 1; i+1 <= n/2; ++i) {
    if(buf[i] == '0') {
      printf("%d %d\n", u, v);
      ++v;
    } else {
      printf("%d %d\n", u, v);
      u = v;
      ++v;
    }
  }
  for(; v <= n; ++v) {
    printf("%d %d\n", u, v);
  }
  
  return 0;
}