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>
#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(1000010, 0);
  int ptr = 0;
  auto check = [&](int dl) {
    for (int i = 0; i < 1000010; ++i) used[i] = 0;
    bool ok = false;
    unique = 0;
    for (int i = 0; i < dl; ++i) {
      if (!used[vec[i]]) ++unique;
      ++used[vec[i]];
    }
    if (unique <= k) ok = true, ptr = 0;
    for (int i = 1; i <= (n - dl); ++i) {
      int hello = vec[i - 1 + dl];
      int goodbye = vec[i - 1];
      used[hello]++;
      unique += (used[hello] == 1);
      used[goodbye]--;
      unique -= (used[goodbye] == 0);
      if (unique <= k) ok = true, ptr = i;
    }
    return ok;
  };
  int l = 1, r = n + 1;
  while (r - l > 1) {
    int mid = l + r >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  check(l);
  cout << ptr + 1 << " " << ptr + l << "\n";
  return 0;
}
