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
length = int(input())
gems = input().split()

result = []
s = 0
dist = set()
pend_e = 0
pend_s = 0
for i in range(0, length):
    cur = gems[i]
    if cur not in dist:
        dist.add(cur)
    else:
        if pend_e != 0:
            result.append([pend_s, pend_e])
        pend_e = i + 1
        pend_s = s + 1
        s = i + 1
        dist.clear()

if s != 0 and pend_e != length + 1:
    result.append([pend_s, length])

if len(result) == 0:
    print(-1)
else:
    print(len(result))

for r in result:
    print(r[0], r[1])
