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
n,k = map(int, raw_input().split())
a = map(int, raw_input().split())
ans,currlen,currstart,ref,count,ansstart,ansend = 0,0,0,{},0,0,0
for i in range(n):
	if a[i] in ref:
		ref[a[i]]+=1
		currlen+=1
	elif count < k:
		ref[a[i]]=1
		currlen+=1
		count+=1
	else:
		if currlen > ans: ans,ansstart,ansend = currlen,currstart,i-1
		while count == k:
			if ref[a[currstart]] == 1:
				del ref[a[currstart]]
				count-=1
			else: ref[a[currstart]] -= 1
			currstart+=1
			currlen-=1
		ref[a[i]]=1
		count+=1
		currlen+=1
if currlen > ans: ans,ansstart,ansend = currlen,currstart,n-1
print ansstart+1, ansend+1
