#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using std::cin;
using std::cout;
using std::queue;
using std::vector;
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
const auto ready = []() {
  cin.tie(0);
  cout.tie(0);
  std::ios_base::sync_with_stdio(false);
  return true;
}();
int main() {
  int n, k;
  cin >> n >> k;
  vi vec(n);
  for (int i = 0; i < n; ++i) cin >> vec[i];
  int ans = 0;
  int unique = 0;
  vi used(1e6 + 10, 0);
  int l = 0, r = 0;
  for (int i = 0; i < n; ++i) {
    while (r < n) {
      if (unique < k || (unique == k && used[vec[r]])) {
        if (!used[vec[r]]) ++unique;
        ++used[vec[r]];
        ++r;
      } else
        break;
    }
    if (unique <= k)
      if (r - i > ans) ans = r - i, l = i;
    --used[vec[i]];
    if (!used[vec[i]]) --unique;
  }
  cout << l + 1 << " " << l + ans << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using std::cin;
using std::cout;
using std::queue;
using std::vector;
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
const auto ready = []() {
  cin.tie(0);
  cout.tie(0);
  std::ios_base::sync_with_stdio(false);
  return true;
}();
int main() {
  int n, k;
  cin >> n >> k;
  vi vec(n);
  for (int i = 0; i < n; ++i) cin >> vec[i];
  int ans = 0;
  int unique = 0;
  vi used(1e6 + 10, 0);
  int l = 0, r = 0;
  for (int i = 0; i < n; ++i) {
    while (r < n) {
      if (unique < k || (unique == k && used[vec[r]])) {
        if (!used[vec[r]]) ++unique;
        ++used[vec[r]];
        ++r;
      } else
        break;
    }
    if (unique <= k)
      if (r - i > ans) ans = r - i, l = i;
    --used[vec[i]];
    if (!used[vec[i]]) --unique;
  }
  cout << l + 1 << " " << l + ans << "\n";
  return 0;
}

```