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
def solve():
    n = input()
    A = map(int,raw_input().split())
    if n == len(set(A)):
        print -1
        return

    ans = []
    l = 0
    r = 0
    i = 0
    visited = {}
    while i < n:
        if A[i] in visited:
            ans.append([l+1,r+1])
            l = r+1
            visited = {}
        else:
            visited[A[i]] = 1
        r += 1
        i+=1
    ans[-1][1] = n
    print len(ans)
    for tup in ans:
        print tup[0], tup[1]

solve()