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 = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

l, r = 0, 0
hsh = [0] * 1000005
idx = [1, 1]
hsh[a[0]] = 1
cnt,ans = 1, -1

while(r < n - 1 and l <= r):
    if hsh[a[r + 1]] == 0 and cnt < k:#when incoming is unique and I can accomodate            
        r += 1
        hsh[a[r]] += 1
        cnt += 1
        if r - l > ans:
            idx = [l + 1, r + 1]
            ans = r - l
        
    elif hsh[a[r + 1]] == 0 and cnt == k:#when I can't accomodate
        while(cnt == k):#look to remove a character from segment
            if hsh[a[l]] == 1:
                cnt -= 1
                hsh[a[l]] -= 1
                l += 1
                break
            else:
                hsh[a[l]] -= 1
                l += 1
        r += 1
        hsh[a[r]] += 1
        cnt += 1
        if r - l > ans:
            idx = [l + 1, r + 1]
            ans = r - l
    elif hsh[a[r + 1]] != 0:
        r += 1
        hsh[a[r]] += 1
        if r - l > ans:
            idx = [l + 1, r + 1]
            ans = r - l
        
print(*idx)
                
            
    



        
        
