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,k = map(int, input().split())
a = list(map(int, input().split()))
# sliding window O(n)
slow,fast = 0,0
cnt = {}
mx = -1
while fast < n:
    x = a[fast]
    cnt[x] = cnt.get(x,0)+1
    if len(cnt) <= k and fast-slow > mx:
        mx = fast-slow
        l,r = slow+1,fast+1
    elif len(cnt) > k:
        x = a[slow]
        cnt[x] -= 1
        if cnt[x] == 0:
            del cnt[x]
        slow += 1
    fast += 1
print(l,r)

# complete search O(n^2)
#dif = -1
#for i in range(n):
#    s = set()
#    for j in range(i,n):
#        s.add(a[j])
#        if len(s) <= k and j-i > dif:
#            dif = j-i
#            l,r = i+1, j+1
#        elif len(s) > k:
#            break;
#print(l,r)
        




