Please note that the memory limit differs from the standard.

You really love to listen to music. During the each of next s days you will listen to exactly m songs from the playlist that consists of exactly n songs. Let's number the songs from the playlist with numbers from 1 to n, inclusive. The quality of song number i is ai.

On the i-th day you choose some integer v (li ≤ v ≤ ri) and listen to songs number v, v + 1, ..., v + m - 1. On the i-th day listening to one song with quality less than qi increases your displeasure by exactly one.

Determine what minimum displeasure you can get on each of the s next days. 

Input

The first line contains two positive integers n, m (1 ≤ m ≤ n ≤ 2·105). The second line contains n positive integers a1, a2, ..., an (0 ≤ ai < 230) — the description of songs from the playlist. 

The next line contains a single number s (1 ≤ s ≤ 2·105) — the number of days that you consider.

The next s lines contain three integers each li, ri, xi (1 ≤ li ≤ ri ≤ n - m + 1; 0 ≤ xi < 230) — the description of the parameters for the i-th day. In order to calculate value qi, you need to use formula: <image>, where ansi is the answer to the problem for day i. Assume that ans0 = 0. 

Output

Print exactly s integers ans1, ans2, ..., anss, where ansi is the minimum displeasure that you can get on day i.

Examples

Input

5 3
1 2 1 2 3
5
1 1 2
1 3 2
1 3 3
1 3 5
1 3 1


Output

2
0
2
3
1
#include <bits/stdc++.h>
using namespace std;
struct answer {
  unsigned key : 20;
  unsigned value : 20;
  answer(unsigned Key = 0, unsigned Value = 0) : key(Key), value(Value) {}
};
inline bool operator<(answer a, answer b) { return a.key < b.key; }
const int offset = 1 << 18;
vector<answer> tree[offset << 1];
int *sum;
pair<int, int> *best;
int *items[offset << 1];
char size[offset << 1];
int n, m;
int pa, pb, pq;
void _insert(int v, int a, int b) {
  if (a >= pb || b <= pa) return;
  if (a >= pa && b <= pb) {
    sum[v]++;
    best[v].first++;
  } else {
    int prev = best[v].second;
    _insert(v * 2, a, (a + b) / 2);
    _insert(v * 2 + 1, (a + b) / 2, b);
    best[v] = min(best[v * 2], best[v * 2 + 1]);
    best[v].first += sum[v];
    if (prev != best[v].second) tree[v].push_back(answer(pq, best[v].second));
  }
}
void insert(int a, int b, int q) {
  pa = a;
  pb = b + 1;
  pq = q;
  _insert(1, 0, offset);
}
int count_sum(int a, int b, int q) {
  a += offset;
  b += offset;
  int ans = 0;
  while (a < b) {
    if (a & 1) {
      ans += lower_bound(items[a], items[a] + (1 << size[a]), q) - items[a];
      a++;
    }
    if ((b & 1) == 0) {
      ans += lower_bound(items[b], items[b] + (1 << size[b]), q) - items[b];
      b--;
    }
    a >>= 1;
    b >>= 1;
  }
  if (a == b)
    ans += lower_bound(items[a], items[a] + (1 << size[a]), q) - items[a];
  return ans;
}
int solve(int a, int b, int q, int pq) {
  int ans = 1000000;
  a += offset;
  b += offset;
  while (a < b) {
    if (a & 1) {
      auto it = upper_bound(tree[a].begin(), tree[a].end(), answer(pq));
      int v = (--it)->value;
      ans = min(ans, count_sum(v, v + m - 1, q));
      a++;
    }
    if ((b & 1) == 0) {
      auto it = upper_bound(tree[b].begin(), tree[b].end(), answer(pq));
      int v = (--it)->value;
      ans = min(ans, count_sum(v, v + m - 1, q));
      b--;
    }
    a >>= 1;
    b >>= 1;
  }
  if (a == b) {
    auto it = upper_bound(tree[a].begin(), tree[a].end(), answer(pq));
    int v = (--it)->value;
    ans = min(ans, count_sum(v, v + m - 1, q));
  }
  return ans;
}
pair<int, int> a[200000];
int nr[200001];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  sum = new int[offset << 1];
  best = new pair<int, int>[offset << 1];
  for (int i = offset; i < offset * 2; i++) {
    best[i].second = i - offset;
    tree[i].push_back(answer(0, best[i].second));
  }
  for (int i = offset - 1; i > 0; i--) {
    best[i].second = best[i * 2].second;
    tree[i].push_back(answer(0, best[i].second));
  }
  nr[0] = 1;
  for (int i = 0; i < n; i++) {
    if (i > 0) {
      nr[i] = nr[i - 1];
      if (a[i].first != a[i - 1].first) nr[i]++;
    }
    insert(max(0, a[i].second - m + 1), a[i].second, nr[i]);
  }
  nr[n] = nr[n - 1] + 1;
  delete[] sum;
  delete[] best;
  for (int i = offset; i < offset * 2; i++) items[i] = new int[1];
  for (int i = 0; i < n; i++) items[a[i].second + offset][0] = a[i].first;
  for (int i = offset - 1; i > 0; i--) {
    size[i] = size[i * 2] + 1;
    int s = 1 << size[i * 2];
    items[i] = new int[s * 2];
    merge(items[i * 2], items[i * 2] + s, items[i * 2 + 1],
          items[i * 2 + 1] + s, items[i]);
  }
  int s, ans = 0;
  scanf("%d", &s);
  while (s--) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    x ^= ans;
    l--;
    r--;
    int px = nr[lower_bound(a, a + n, make_pair(x, 0)) - a] - 1;
    ans = solve(l, r, x, px);
    printf("%d\n", ans);
  }
}
