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

def main():
    n,k = map(int, input().split())
    myList = list(map(int, input().split()))
    #n,k = 5,5
    #myList = [1,2,3,4,5]
    i = 0
    distinct = 0
    dictOf = [0]*1000001
    best = 0
    bestX = 0
    bestY = 0
    for j in range(n):
        dictOf[myList[j]] += 1
        if dictOf[myList[j]] == 1:
            distinct += 1
        while distinct > k:
            dictOf[myList[i]] -= 1
            if dictOf[myList[i]] == 0:
                distinct -= 1
            i += 1
        if j-i+1 > best:
            best = j-i+1
            bestX = i+1
            bestY = j+1
    print(bestX, bestY)

if __name__ == '__main__':
    main()
    

