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
import sys
n, m = map(int, sys.stdin.readline().split())
vis = [0] * 1000005
num = list(map(int, sys.stdin.readline().split()))
flag = 0
temp = num[0]
for i in range(n):
    if num[i] != temp:
        flag = 1
        break
if flag == 0:
    print("{0} {1}".format(1, n))
else:
    l = 0
    r = 0
    al = 0
    ar = 0
    ans = 0
    now = 0
    for i in range(n):
        vis[num[i]] += 1
        if vis[num[i]] == 1:
            now += 1
        while now > m:
            vis[num[l]] -= 1
            if vis[num[l]] == 0:
                now -= 1
            l += 1
        if i - l + 1 > ar - al + 1:
            ar = i
            al = l
    print('{0} {1}'.format(al+1, ar+1))





