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
var n = readline();
var pearls = readline().split(' ');

var counter = [];

var result = [];
var rLength = 0;

var i;
var start_idx = 0;

pearls.map(function(x, idx) {
    x = parseInt(x, 10);
    if(!counter[x]) {
        counter[x] = 1;
    } else {
        result[rLength] = [start_idx + 1, idx +1];
        start_idx = idx + 1;
        rLength = rLength + 1;
        counter = [];
    }
});

if (rLength == 0) {
    print("-1");
}
else {
    result[rLength-1][1] = parseInt(n, 10);

    print(rLength);
    for (i = 0; i < rLength; i++){
        print(result[i][0]+ ' ' + result[i][1]);
    }
}