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
#!/bin/python3

import sys
n = int(input())
a = list(map(int, input().split()))

ans = [];
sa = sorted(a);
cmpr = {};
for i in range(0, len(a)):
    cmpr[sa[i]] = i;
cid = [0 for i in range(0,n)];

id = 1;
pi = 0
for i in range(0,n):
    if cid[cmpr[a[i]]] == id:
        ans.append(i + 1)
        id+=1
        continue;
    cid[cmpr[a[i]]] = id
if len(ans) == 0:
    print(-1)
else:
    ans[-1] = n;
    print(len(ans))
    ans = [0] + ans
    for i in range(0, len(ans) - 1):
        print(ans[i] + 1 , ans[i + 1] )
