Problem: 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

Solution (in PYTHON3):

Solution:
n = input()
p = input()
n = n.split(' ')
k = int(n[1])
n = int(n[0])
p = p.split(' ')
p = [int(q) for q in p]
cnt = [0 for i in range(0,int(1e6 + 1))]
curcnt = 0
l = 0
r = -1
maxlen = -1
maxl = 0
maxr = 0
while r<n-1 and l<=n:
    while(r<n-1 and (curcnt + (cnt[p[r+1]] == 0)) <=k):
        curcnt = curcnt + (cnt[p[r+1]] == 0)
        cnt[p[r+1]] +=1 
        r+=1
    if maxlen <(r-l):
        maxlen = r-l
        maxr = r
        maxl = l
    cnt[p[l]]-=1
    curcnt-=(cnt[p[l]]==0)
    l+=1
print(str(maxl+1)+' '+str(maxr+1))

#16 3
#4 2 6 9 6 8 3 3 4 8 2 3 4 6 9 6
#13 3
#1 2 3 4 3 5 6 6 1 3 4 3 7
#16 3
#1 2 6 9 6 8 3 3 1 8 2 3 6 9 6