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;
const int INF = 1e9;
const long long LINF = 1e18;
const long double EPS = 1e-10;
const long double PI = 3.141592653589;
set<int> cnt;
int main() {
  ios::sync_with_stdio(false);
  float tbegin = clock();
  srand(42);
  if ("" != "") {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  int n;
  scanf("%d\n", &n);
  int last = -1;
  vector<pair<int, int>> ans;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    if (cnt.count(a)) {
      ans.push_back(make_pair(last + 1, i));
      last = i;
      cnt.clear();
    } else {
      cnt.insert(a);
    }
  }
  if (last == -1) {
    printf("-1\n");
  } else {
    printf("%d\n", (int)ans.size());
    if (last != n - 1) {
      ans.rbegin()->second = n - 1;
    }
    for (auto it : ans) {
      printf("%d %d\n", it.first + 1, it.second + 1);
    }
  }
  return 0;
}
