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.

Input

The first line contains three space-separated integers n, k, d (1 ≤ n ≤ 2·105; 0 ≤ k ≤ 2·105; 0 ≤ d ≤ 109). The second line contains n space-separated integers: a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the actual sequence.

Output

Print two space-separated integers l, r (1 ≤ l ≤ r ≤ n) show that sequence al, al + 1, ..., ar is the longest subsegment that is a good k-d sequence.

If there are multiple optimal answers, print the one with the minimum value of l.

Examples

Input

6 1 2
4 3 2 8 6 2


Output

3 5

Note

In the first test sample the answer is the subsegment consisting of numbers 2, 8, 6 — after adding number 4 and sorting it becomes sequence 2, 4, 6, 8 — the arithmetic progression with difference 2.
#include <bits/stdc++.h>
using namespace std;
int rd() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') f = c == '-' ? -1 : 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * f;
}
const int N = 3e5 + 10;
const int inf = 1e6 + 10;
map<int, int> mp;
int a[N], s1[N], s2[N], sum[N << 2], tag[N << 2];
int L, R, ans, n, K, d, top1, top2, k;
void modify(int k, int l, int r, int x, int y, int z) {
  if (x == l && r == y) {
    sum[k] += z, tag[k] += z;
    return;
  }
  int mid = l + r >> 1;
  if (y <= mid)
    modify(k << 1, l, mid, x, y, z);
  else if (x > mid)
    modify(k << 1 | 1, mid + 1, r, x, y, z);
  else
    modify(k << 1, l, mid, x, mid, z),
        modify(k << 1 | 1, mid + 1, r, mid + 1, y, z);
  tag[k] = min(tag[k << 1], tag[k << 1 | 1]) + sum[k];
}
int query(int k, int l, int r, int x) {
  if (l == r) return l;
  int mid = l + r >> 1;
  x -= sum[k];
  if (tag[k << 1] <= x)
    return query(k << 1, l, mid, x);
  else
    return query(k << 1 | 1, mid + 1, r, x);
}
int main() {
  n = rd(), K = rd(), d = rd();
  for (int i = 1; i <= n; i++) a[i] = rd() + 1e9 + 1;
  L = R = ans = 1;
  if (!d) {
    for (int j, i = 1; i <= n; i = j + 1) {
      for (j = i; j < n && a[j + 1] == a[i]; j++)
        ;
      if (j - i + 1 > ans) ans = j - i + 1, L = i, R = j;
    }
    return printf("%d %d\n", L, R), 0;
  }
  modify(1, 1, n, 1, n, inf);
  for (int i = 1; i <= n; i++) {
    if (i != 1 && (a[i] - a[i - 1]) % d)
      modify(1, 1, n, k + 1, i - 1, inf), k = i - 1;
    if (mp[a[i]] > k) modify(1, 1, n, k + 1, mp[a[i]], inf), k = mp[a[i]];
    mp[a[i]] = i;
    while (top1 && a[s1[top1]] > a[i])
      modify(1, 1, n, s1[top1 - 1] + 1, s1[top1], a[s1[top1]] / d - a[i] / d),
          top1--;
    while (top2 && a[s2[top2]] < a[i])
      modify(1, 1, n, s2[top2 - 1] + 1, s2[top2], a[i] / d - a[s2[top2]] / d),
          top2--;
    s1[++top1] = s2[++top2] = i;
    modify(1, 1, n, i, i, i - inf);
    int j = query(1, 1, n, i + K);
    if (i + L > R + j) L = j, R = i;
  }
  return printf("%d %d\n", L, R), 0;
}
