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, input().split())
a = list(map(int, input().split()))

cnt = [0] * 1000001
t = 0

left = 0
right = -1

l = 0
r = 0
while r < n:
    if cnt[a[r]] > 0: # already have this one
        cnt[a[r]] += 1
        r += 1
    elif t < k: # still has capacity
        cnt[a[r]] = 1
        t += 1
        r += 1
    else:
        # final answer for starting from l
        if r - l > right - left + 1:
            left = l
            right = r - 1

        # prepare for next answer
        cnt[a[l]] -= 1
        if cnt[a[l]] == 0:
            t -= 1
        l += 1

if r - l > right - left + 1:
    left = l
    right = r - 1

print(left + 1, right + 1)