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
#include <bits/stdc++.h>
using namespace std;
int a[300005], c1[300005], c2[300005];
map<int, int> b;
int main() {
  long i, j, k, m, n, x, y, m1, m2, p, q, x1, x2, y1, y2;
  long long sum;
  bool flag;
  while (cin >> n) {
    k = 0;
    x = 1;
    b.clear();
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (b[a[i]]) {
        c1[k] = x;
        c2[k++] = i;
        x = i + 1;
        b.clear();
      } else
        b[a[i]] = 1;
    }
    c2[k - 1] = n;
    if (k)
      cout << k << endl;
    else
      cout << "-1" << endl;
    for (i = 0; i < k; i++) cout << c1[i] << ' ' << c2[i] << endl;
  }
  return 0;
}
