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;
long long power(long long x, long long y, long long p);
int Bit(long long x);
long long Sum(int x);
inline int D() {
  int t;
  scanf("%d", &t);
  return t;
}
inline long long llD() {
  long long t;
  scanf("%lld", &t);
  return t;
}
const int N = 1e6 + 5;
int arr[N], fr[N];
int l, r, mid, n, k, mn, mx, def, h;
bool can(int idx) {
  memset(fr, 0, sizeof(fr));
  def = 0;
  for (int i = 1; i <= idx; ++i) def += (fr[arr[i]] == 0), ++fr[arr[i]];
  h = def;
  if (def <= k) mn = 1, mx = idx;
  for (int i = idx + 1, j = 1; i <= n; ++i, ++j) {
    def += (fr[arr[i]] == 0);
    ++fr[arr[i]], --fr[arr[j]];
    def -= (fr[arr[j]] == 0);
    h = min(h, def);
    if (def <= k) mn = j + 1, mx = i;
  }
  return h <= k;
}
void BS() {
  l = 0, r = n, mid;
  while (l < r) {
    mid = (l + r + 1) >> 1;
    if (can(mid))
      l = mid;
    else
      r = mid - 1;
  }
  printf("%d %d", mn, mx);
}
int main() {
  n = D(), k = D();
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
  BS();
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int Bit(long long x) {
  if (!x) return 0;
  return 1 + Bit(x >> 1);
}
long long Sum(int x) { return (1ll * x * (x + 1)) >> 1ll; }