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
import sys
import math
from collections import defaultdict
from bisect import bisect_left, bisect_right

sys.setrecursionlimit(10**7)
def input():
    return sys.stdin.readline()[:-1]

mod = 10**9 + 7

def I(): return int(input())
def LI(): return list(map(int, input().split()))
def LIR(row,col):
    if row <= 0:
        return [[] for _ in range(col)]
    elif col == 1:
        return [I() for _ in range(row)]
    else:
        read_all = [LI() for _ in range(row)]
        return map(list, zip(*read_all))

#################

s = list(input())
n = len(s)

for i in range((n-1)//2+1):
    if s[i] != s[n-2-i]:
        print(-1)
        exit()

if s[0] == '0':
    print(-1)
    exit()

if s[n-1] == '1':
    print(-1)
    exit()

from_ = 1
to_ = 2
ans = []
for i in range(n-1):
    ans.append((from_,to_))
    if s[i] == '1':
        from_ = to_
        to_ += 1
    else:
        to_ += 1

for a in ans:
    print(*a)