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
count = 0
N=int(input())
A=list(map(int,input().split()))+['-1']
D={}
Ans=[]
a,b=1,1
for val in A:
    if val in D:
        D[val] += 1
    else:
        D[val] = 1
    if D[val] % 2 ==0:
        Ans.append([str(a),str(b)])
        a=b+1
        D={}
    b += 1
L=len(Ans)
if L ==0:
    print(-1)
else:
    Ans[-1][1]=str(N)
    print(L)
    for val in Ans:
        print(' '.join(val))
    