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
s = input().split()
n, k = int(s[0]), int(s[1])
s = input().split()
a = []
for i in range(n):
    e = int(s[i])
    a.append(e)
freq, add = [], []
for i in range(10 ** 6 + 1):
    freq.append(0)
for i in range(n):
    add.append(1)
i, j = 0, 0
d = 0
m = -1
l, r = -1, -1
while i <= j and j < n:
    d += (freq[a[j]] == 0)
    freq[a[j]] += add[j]
    add[j] = 0
    if d > k:
        freq[a[i]] = max(0, freq[a[i]] - 1)
        d -= (freq[a[i]] == 0)
        i += 1
    else:
        temp = j - i
        if temp > m:
            m = temp
            l, r = i + 1, j + 1
        j += 1
print(l, r)