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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
long long c[1000005];
long long a[500005];
int main() {
  long long x, y, ans = 0, p = 1, i, j, k, l, m, n, kk = 0;
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (c[a[i]] == 0) kk++;
    c[a[i]]++;
    if (kk > k) {
      while (1) {
        c[a[p]]--;
        p++;
        if (c[a[p - 1]] == 0) break;
      }
      kk = k;
    }
    if (ans < i - p + 1) {
      ans = i - p + 1;
      x = p;
      y = i;
    }
  }
  cout << x << " " << y;
}
