#include <bits/stdc++.h>
using namespace std;
int hap[100010];
int rec[100010];
int dif[100010];
int main() {
  bool flag;
  int n, k, tol, ans, min, max;
  while (cin >> n >> k) {
    flag = false;
    tol = 0;
    memset(dif, -1, sizeof(dif));
    for (int i = 1; i <= n; i++) scanf("%d", &rec[i]);
    for (int i = 1; i <= n; i++) {
      if (dif[rec[i]] != -1) {
        dif[rec[i]] = i;
      }
      if (dif[rec[i]] == -1) {
        tol++;
        hap[tol] = rec[i];
        dif[rec[i]] = i;
      }
      if (tol == k) {
        flag = true;
        ans = i;
        break;
      }
    }
    if (flag) {
      max = -1;
      min = 100010;
      for (int i = 1; i <= k; i++) {
        if (dif[hap[i]] < min) min = dif[hap[i]];
        if (dif[hap[i]] > max) max = dif[hap[i]];
      }
      printf("%d %d\n", min, max);
    } else {
      printf("-1 -1\n");
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You've got an array a, consisting of n integers: a1, a2, ..., an. Your task is to find a minimal by inclusion segment [l, r] (1 ≤ l ≤ r ≤ n) such, that among numbers al, al + 1, ..., ar there are exactly k distinct numbers.

Segment [l, r] (1 ≤ l ≤ r ≤ n; l, r are integers) of length m = r - l + 1, satisfying the given property, is called minimal by inclusion, if there is no segment [x, y] satisfying the property and less then m in length, such that 1 ≤ l ≤ x ≤ y ≤ r ≤ n. Note that the segment [l, r] doesn't have to be minimal in length among all segments, satisfying the given property.

Input

The first line contains two space-separated integers: n and k (1 ≤ n, k ≤ 105). The second line contains n space-separated integers a1, a2, ..., an — elements of the array a (1 ≤ ai ≤ 105).

Output

Print a space-separated pair of integers l and r (1 ≤ l ≤ r ≤ n) such, that the segment [l, r] is the answer to the problem. If the sought segment does not exist, print "-1 -1" without the quotes. If there are multiple correct answers, print any of them.

Examples

Input

4 2
1 2 2 3


Output

1 2


Input

8 3
1 1 2 2 3 3 4 5


Output

2 5


Input

7 4
4 7 7 4 7 4 7


Output

-1 -1

Note

In the first sample among numbers a1 and a2 there are exactly two distinct numbers.

In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.

In the third sample there is no segment with four distinct numbers.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int hap[100010];
int rec[100010];
int dif[100010];
int main() {
  bool flag;
  int n, k, tol, ans, min, max;
  while (cin >> n >> k) {
    flag = false;
    tol = 0;
    memset(dif, -1, sizeof(dif));
    for (int i = 1; i <= n; i++) scanf("%d", &rec[i]);
    for (int i = 1; i <= n; i++) {
      if (dif[rec[i]] != -1) {
        dif[rec[i]] = i;
      }
      if (dif[rec[i]] == -1) {
        tol++;
        hap[tol] = rec[i];
        dif[rec[i]] = i;
      }
      if (tol == k) {
        flag = true;
        ans = i;
        break;
      }
    }
    if (flag) {
      max = -1;
      min = 100010;
      for (int i = 1; i <= k; i++) {
        if (dif[hap[i]] < min) min = dif[hap[i]];
        if (dif[hap[i]] > max) max = dif[hap[i]];
      }
      printf("%d %d\n", min, max);
    } else {
      printf("-1 -1\n");
    }
  }
  return 0;
}

```