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 maxn = 1e6 + 7, maxd = 1e4 + 7, modd = 1e9 + 7;
const long long INF = 0x7f7f7f7f;
long long n;
set<long long> st;
struct edge {
  long long l_, r_;
} e[maxn];
signed main() {
  scanf("%lld", &n);
  long long cnt = 0;
  for (register long long i = 1; i <= n; ++i) {
    long long x;
    scanf("%lld", &x);
    if (st.empty()) {
      st.insert(x);
      e[cnt].l_ = i;
    } else {
      if (st.count(x)) {
        st.clear();
        e[cnt++].r_ = i;
      } else {
        st.insert(x);
      }
    }
  }
  if (!cnt) {
    puts("-1");
  } else {
    e[cnt - 1].r_ = n;
    printf("%lld\n", cnt);
    for (register long long i = 0; i < cnt; ++i) {
      printf("%lld %lld\n", e[i].l_, e[i].r_);
    }
  }
  return 0;
}