#include<bits/stdc++.h>
using namespace std;

const int N = 100005;
int n;
char a[N];
vector<int> v;

int main()
{
	scanf("%s",a+1);
	n = strlen(a+1);
	if(a[1] == '0' || a[n] == '1') {
		puts("-1");
		return 0;
	}
	for(int i=1;i<=n-1;i++) {
		if(a[i] != a[n-i]) {
			puts("-1");
			return 0;
		}
	}
	for(int i=1;i<=n-1;i++) {
		if(a[i] == '1') v.push_back(i);
	}
	v.push_back(n);
	for(int i=1;i<(int)v.size();i++) {
		printf("%d %d\n",v[i-1],v[i]);
		for(int j=v[i-1]+1;j<v[i];j++) {
			printf("%d %d\n",j,v[i]);
		}
	}
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

Examples

Input

1111


Output

-1


Input

1110


Output

1 2
2 3
3 4


Input

1010


Output

1 2
1 3
1 4

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int N = 100005;
int n;
char a[N];
vector<int> v;

int main()
{
	scanf("%s",a+1);
	n = strlen(a+1);
	if(a[1] == '0' || a[n] == '1') {
		puts("-1");
		return 0;
	}
	for(int i=1;i<=n-1;i++) {
		if(a[i] != a[n-i]) {
			puts("-1");
			return 0;
		}
	}
	for(int i=1;i<=n-1;i++) {
		if(a[i] == '1') v.push_back(i);
	}
	v.push_back(n);
	for(int i=1;i<(int)v.size();i++) {
		printf("%d %d\n",v[i-1],v[i]);
		for(int j=v[i-1]+1;j<v[i];j++) {
			printf("%d %d\n",j,v[i]);
		}
	}
}

```