The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1
#!/usr/bin/env python

n = int(raw_input())
a = map(int, raw_input().split())

segments = []
start = 1
nrs = set([])
idx = 1

for p in a:
  if p in nrs:
    segments.append((start,idx))
    start = idx+1
    nrs = set([])
  else:
    nrs.add(p)
  idx+=1

if len(segments) == 0:
  print -1
else:
  print len(segments)
  segments[-1]=(segments[-1][0], idx-1)
  for s in segments:
    print "%d %d" % (s[0], s[1])
