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 = int(input())
a = [k for k in (input()).split(' ')]
answer = -1
l = []
dict = {}
i = 0
for j in range(len(a)):
    if dict.get(a[j]):
        dict = {}
        answer+=1
        l.append(i)
        l.append(j)
        i = j + 1
    else:
        dict.update({a[j] : 1})
if len(l)>=1:
    l[len(l) - 1] = n-1
if answer == -1:
    print(answer)
else:
    print(int(answer + 1))
    for p in l:
        print(str(p + 1) + ' ')