We'll call a sequence of integers a good *k*-*d* sequence if we can add to it at most *k* numbers in such a way that after the sorting the sequence will be an arithmetic progression with difference *d*.

You got hold of some sequence *a*, consisting of *n* integers. Your task is to find its longest contiguous subsegment, such that it is a good *k*-*d* sequence.
The first line contains three space-separated integers *n*, *k*, *d* (1 ≤ *n* ≤ 2·10<sup class="upper-index">5</sup>; 0 ≤ *k* ≤ 2·10<sup class="upper-index">5</sup>; 0 ≤ *d* ≤ 10<sup class="upper-index">9</sup>). The second line contains *n* space-separated integers: *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> ( - 10<sup class="upper-index">9</sup> ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>) — the actual sequence.
n, k, d = map(int, input().split())
a = list(map(int, input().split()))

best_l, best_r = 1, 1
l, r = 1, 1
curr_diff = 0
count = [0] * (2 * 10**5 + 1)

while r <= n:
    count[a[r - 1]] += 1
    if count[a[r - 1]] == 1:
        curr_diff += 1

    while curr_diff > k:
        count[a[l - 1]] -= 1
        if count[a[l - 1]] == 0:
            curr_diff -= 1
        l += 1

    if (a[r - 1] - a[l - 1]) / (r - l) == d and r - l + 1 > best_r - best_l + 1:
        best_l, best_r = l, r

    r += 1

print(best_l, best_r)
