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
size = input()
myDic,capture,limit = {},[],-1
for x,y in enumerate(map(int,raw_input().split())):
    if y in myDic:
        check = x - myDic[y][-1]
        if myDic[y][-1] > limit:
            capture.append((myDic[y][-1]+1,x+1))
            limit = x; del myDic[y]
        else: myDic[y][-1] = x
    else:
        myDic.update({y:[x]})
if capture:
    res = len(capture)
    if res == 1:
        print res
        print 1,size
    else:
        print res
        print 1,capture[0][1]
        start = capture[0][1]+1
        for c in range(1,res):
            print start,capture[c][1] if c + 1 < res else size 
            start = capture[c][1] + 1
else: print -1