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
import sys
def i(): return sys.stdin.readline().strip()
sys.setrecursionlimit(99999999)
ii=lambda:map(int,i().split(" "))

n=int(i())
ll=map(int,i().split(" "))
back=[-1]*n
currentSegment={}
last=0
for e in xrange(n):
    current=ll[e]
    if current in currentSegment:
        back[e]=last
        last=e+1
        currentSegment={}
    else:
        currentSegment[current]=True
acc=[]
for e in xrange(n):
    if back[e]<> -1:
        acc.append((back[e],e))
#print acc
if acc==[]:
    print -1
else:
    acc[-1]=(acc[-1][0],n-1)
    print len(acc)
    for e in acc:
        a,b=e
        print a+1,b+1