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;
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const double PI = acos(-1.0);
const long long mod = 1000000007;
const long long INF = 1e9;
const long long LINF = 1e18;
const long long MAX = 2e5 + 7;
const long long MAXN = 5e5;
const long long MAXA = 1e6 + 1;
const double EPS = 1e-9;
int n;
int k;
int a[MAXN];
int cnt[MAXA];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = (0); i < (n); ++i) cin >> a[i];
  int diff = 0;
  int l = 0, r = 0;
  pair<int, int> ans = make_pair(0, 0);
  int best = 1;
  while (l < n) {
    while (r < n) {
      if (!cnt[a[r]]) ++diff;
      if (diff > k) {
        --diff;
        if (r - l > best) {
          ans = make_pair(l, r - 1);
          best = r - l;
        }
        break;
      } else
        cnt[a[r++]]++;
    }
    if (r == n && diff <= k) {
      if (r - l > best) {
        best = r - l;
        ans = make_pair(l, r - 1);
      }
      break;
    }
    cnt[a[l]]--;
    if (!cnt[a[l]]) --diff;
    ++l;
    r = max(l, r);
  }
  cout << ans.first + 1 << " " << ans.second + 1;
  cin >> n;
  return 0;
}
