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;
const int N = 2e5 + 7;
const int SIZE = 400;
int n, m;
int a[N], sum[N], look[N];
struct Block {
  vector<int> pos;
  vector<int> best;
  vector<int> first_v;
  vector<pair<int, pair<int, int> > > keys;
  Block() {
    pos.clear();
    best.clear();
    first_v.clear();
    keys.clear();
  }
  void init() {
    sort(pos.begin(), pos.end());
    pos.erase(unique(pos.begin(), pos.end()), pos.end());
    while (pos.size() && pos.back() >= n) pos.pop_back();
    pos.push_back(n);
    for (int i = 0, pre = 0; i + 1 < pos.size(); i++) {
      int max_v = -1;
      first_v.push_back(pre + sum[pos[i]]);
      for (int j = pos[i]; j < pos[i + 1]; j++) {
        max_v = max(max_v, pre += sum[j]);
      }
      best.push_back(max_v);
    }
  }
  void insert_key(int p, int v, int delta) {
    p = lower_bound(pos.begin(), pos.end(), p) - pos.begin();
    keys.push_back(make_pair(p, make_pair(v, delta)));
  }
  void init_key() { sort(keys.begin(), keys.end()); }
  int query(int l, int r, int v) {
    int ans = 0, delta = 0;
    for (int i = 0, ptr = 0; i + 1 < pos.size(); i++) {
      while (ptr < keys.size() && keys[ptr].first <= i) {
        delta += (keys[ptr].second.first >= v) * keys[ptr].second.second;
        ptr++;
      }
      if (r < pos[i] || pos[i + 1] < l) {
        continue;
      } else if (l <= pos[i] && pos[i + 1] <= r) {
        ans = max(ans, delta + best[i]);
      } else {
        for (int j = pos[i], pre = first_v[i] + delta; j < pos[i + 1] && j <= r;
             j++) {
          if (l <= j && j <= r) {
            ans = max(ans, pre);
          }
          pre -= (a[j] >= v);
          pre += (a[j + m] >= v);
        }
      }
    }
    return ans;
  }
};
vector<Block> blocks;
int main() {
  vector<pair<int, int> > pairs;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pairs.push_back(make_pair(a[i], i));
  }
  sort(pairs.begin(), pairs.end());
  for (int i = n - 1; i >= 0; i--) {
    if (i % SIZE == 0) {
      Block b;
      for (int j = i; j < i + SIZE && j < n; j++) {
        look[j] = blocks.size();
        int l = max(0, pairs[j].second - m + 1), r = pairs[j].second + 1;
        b.pos.push_back(l);
        b.pos.push_back(r);
      }
      for (int k = 0; k < n; k += SIZE) {
        b.pos.push_back(k);
      }
      b.init();
      for (int j = i; j < i + SIZE && j < n; j++) {
        int l = max(0, pairs[j].second - m + 1), r = pairs[j].second + 1;
        b.insert_key(l, pairs[j].first, 1);
        b.insert_key(r, pairs[j].first, -1);
        sum[l]++;
        sum[r]--;
      }
      b.init_key();
      blocks.push_back(b);
    }
  }
  int q, ans = 0;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    l--, r--;
    x ^= ans;
    int t = lower_bound(pairs.begin(), pairs.end(), make_pair(x, -1)) -
            pairs.begin();
    printf("%d\n", ans = m - blocks[look[t]].query(l, r, x));
  }
  return 0;
}
