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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long n, k;
const long long maxn = 5e5 + 10;
long long a[maxn];
const long long inf = 1e15;
pair<long long, pair<long long, long long> > check(long long mid) {
  pair<long long, pair<long long, long long> > ans = {0, {-1, -1}};
  vector<long long> freq(1e6 + 10, 0);
  long long cnt = 0;
  for (long long i = 1; i <= mid; ++i) {
    if (!freq[a[i]]) ++cnt;
    ++freq[a[i]];
  }
  if (cnt <= k) {
    ans = {1, {1, mid}};
    return ans;
  }
  for (long long i = mid + 1; i <= n; ++i) {
    long long prev = i - mid;
    --freq[a[prev]];
    if (!freq[a[prev]]) --cnt;
    if (!freq[a[i]]) ++cnt;
    ++freq[a[i]];
    if (cnt <= k) {
      ans = {1, {prev + 1, i}};
      return ans;
    }
  }
  return ans;
}
pair<long long, long long> binsearch(long long lo, long long hi) {
  long long len = -inf;
  pair<long long, long long> ans = {-1, -1};
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    pair<long long, pair<long long, long long> > cur = check(mid);
    if (cur.first) {
      lo = mid + 1;
      if (mid > len) {
        len = mid;
        ans = cur.second;
      }
    } else {
      hi = mid - 1;
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(9);
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  pair<long long, long long> ans = binsearch(1, n);
  cout << ans.first << " " << ans.second << '\n';
  return 0;
}