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 pearlsAmount = readline();
var typesOfPearls = readline().split(' ');
var segmentsEndPoints = [];
var amountOfSegments = 0;
var typesExistence = [];
for (var i = 0; i < pearlsAmount; i++) {
    if (typesExistence[typesOfPearls[i]]) {
        amountOfSegments++;
        segmentsEndPoints.push(i+1);
        typesExistence = [];
    } else {
        typesExistence[typesOfPearls[i]] = 1;
    }
}

if (amountOfSegments == 0) {
    print("-1");
} else {
    segmentsEndPoints.pop();
    segmentsEndPoints.push(pearlsAmount);
    print(amountOfSegments);
    var startPoint = 1;
    for (var i = 0; i < amountOfSegments; i++) {
        print(startPoint + " " + segmentsEndPoints[i]);
        startPoint = segmentsEndPoints[i] + 1;
    }
}