Problem: 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

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout
n, k = [int(i) for i in stdin.readline().split()]
s = [0] * (10**6 + 1)
a = [int(i) for i in stdin.readline().split()]
mx = lmx = rmx = l = sm = 0
r = -1
while True:
    r += 1
    if r == n:
        mx, lmx, rmx = max((mx, lmx, rmx), (r - l, l, r - 1))
        break
    if not s[a[r]]:
        sm += 1
    s[a[r]] += 1
    if sm > k:
        mx, lmx, rmx = max((mx, lmx, rmx), (r - l, l, r - 1))
        while sm > k and l < r:
            s[a[l]] -= 1
            if not s[a[l]]:
                sm -= 1
            l += 1
    elif r == n - 1:
        mx, lmx, rmx = max((mx, lmx, rmx), (r - l + 1, l, r))
        break
print(lmx + 1, rmx + 1)