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
R = lambda: map(int, input().split())
n, k = R()
arr = list(R())
rl, rr = 0, 0
l = 0
rec = {}
for r, v in enumerate(arr):
    rec.setdefault(v, 0)
    rec[v] += 1
    while len(rec) > k:
        rec[arr[l]] -= 1
        if not rec[arr[l]]:
            rec.pop(arr[l])
        l += 1
    if rr - rl < r - l:
        rl, rr = l, r
print(rl + 1, rr + 1)