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()))
ans = 0
cur_r = 0
d = {}
cur_ans = 0
ans_1 = 0
ans_2 = 0

for cur_l in range(n):
    t = a[cur_l]
    d[t] = d.get(t, 0)+1
    if d[t] == 1:
        cur_ans += 1

    while cur_ans > k:
        t = a[cur_r]
        d[t] -= 1
        if not(d[t]):
            cur_ans -= 1
        cur_r += 1

    if cur_l-cur_r+1 > ans:
        ans = cur_l-cur_r+1
        ans_1 = cur_r
        ans_2 = cur_l

print(ans_1+1, ans_2+1)
