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 = int(input())
ty = list(map(int, input().split()))
num = 0
x = 1
seg = []
a = 1
dic = {}
for i in range(n):
    dic[ty[i]] = 0
dic[ty[0]] = 1
for i in range(1, n):
    jud = 0
    if dic[ty[i]] != num + 1:
        dic[ty[i]] = num + 1
    elif dic[ty[i]] == num + 1:
        jud = 1
        num += 1
        b = i + 1
        ls = [str(a), str(b)]
        seg.append(ls)
        a = i + 2
if len(seg) == 0:
    print(-1)
else:
    if jud == 0:
        seg[-1][1] = str(n)
    print(num)
    for i in range(len(seg)):
        print(' '.join(seg[i]))




