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())
cnt = 0
gems = list(map(int, input().split()))
pearls = set()
for i in range(n):
    if gems[i] not in pearls:
        pearls.add(gems[i])
    else:
        cnt += 1
        pearls = set()
    
if cnt:
    print(cnt)
    first = 0
    second = 0
    pearls = set()
    for i in range(n):
        if gems[i] not in pearls:
            pearls.add(gems[i])
        else:
            if second:
                print(first + 1, second + 1)
                first = second + 1
            second = i
            pearls = set()
    print(first + 1, n)
else:
    print('-1')