#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
const int B = 512;
int song[N], order[N], quality[N], lazy[B], maxi[B];
int lazy_history[B][B], maxi_history[B][B];
vector<array<int, 3>> partial[N], updates[N];
vector<array<int, 2>> whole[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> song[i];
  }
  iota(order, order + n, 0);
  sort(order, order + n, [&](int a, int b) { return song[a] > song[b]; });
  for (int i = 0; i < n; ++i) {
    if (i % B == 0) {
      for (int j = 0; j <= (n - 1) / B; ++j) {
        lazy_history[i / B][j] = lazy[j];
        maxi_history[i / B][j] = maxi[j];
      }
    }
    int r = order[i], l = max(0, r - m + 1);
    if (l / B == r / B) {
      partial[i].push_back({l, r, 0});
    } else {
      if (l % B != 0) {
        int a = l - l % B + B;
        partial[i].push_back({l, a - 1, 0});
        l = a;
      }
      if (r % B != B - 1) {
        int b = r - r % B - 1;
        partial[i].push_back({b + 1, r, 0});
        r = b;
      }
      whole[i].push_back({l / B, r / B});
    }
    for (auto &[a, b, x] : partial[i]) {
      for (int j = a; j <= b; ++j) {
        maxi[j / B] = max(maxi[j / B], ++quality[j]);
      }
      x = maxi[a / B];
      updates[a / B].push_back({a % B, b % B, i});
    }
    for (auto [a, b] : whole[i]) {
      for (int j = a; j <= b; ++j) {
        ++lazy[j];
      }
    }
  }
  int s, ans = 0;
  cin >> s;
  while (s--) {
    int l, r, q;
    cin >> l >> r >> q;
    --l, --r;
    q = lower_bound(order, order + n, q ^ ans,
                    [&](int a, int x) { return x <= song[a]; }) -
        order;
    fill(lazy, lazy + (n - 1) / B + 1, 0);
    for (int i = 0; i <= (n - 1) / B; ++i) {
      maxi[i] = maxi_history[q / B][i];
      lazy[i] += lazy_history[q / B][i];
      lazy[i + 1] -= lazy_history[q / B][i];
    }
    for (int i = q - q % B; i < q; ++i) {
      for (auto [a, b, c] : partial[i]) {
        maxi[a / B] = c;
      }
      for (auto [a, b] : whole[i]) {
        ++lazy[a];
        --lazy[b + 1];
      }
    }
    for (int i = 1; i <= (n - 1) / B; ++i) {
      lazy[i] += lazy[i - 1];
    }
    ans = 0;
    vector<array<int, 2>> segments;
    if (l / B == r / B) {
      segments.push_back({l, r});
    } else {
      int a = l - l % B + B, b = r - r % B - 1;
      segments.push_back({l, a - 1});
      segments.push_back({b + 1, r});
      for (int i = a / B; i <= b / B; ++i) {
        ans = max(ans, lazy[i] + maxi[i]);
      }
    }
    for (auto [u, v] : segments) {
      fill(maxi, maxi + B, 0);
      for (auto [a, b, i] : updates[u / B]) {
        if (i < q) {
          ++maxi[a];
          if (b + 1 < B) {
            --maxi[b + 1];
          }
        }
      }
      for (int i = 0; i < B; ++i) {
        if (i > 0) {
          maxi[i] += maxi[i - 1];
        }
        if (i >= u % B && i <= v % B) {
          ans = max(ans, lazy[u / B] + maxi[i]);
        }
      }
    }
    ans = m - ans;
    cout << ans << "\n";
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
const int B = 512;
int song[N], order[N], quality[N], lazy[B], maxi[B];
int lazy_history[B][B], maxi_history[B][B];
vector<array<int, 3>> partial[N], updates[N];
vector<array<int, 2>> whole[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> song[i];
  }
  iota(order, order + n, 0);
  sort(order, order + n, [&](int a, int b) { return song[a] > song[b]; });
  for (int i = 0; i < n; ++i) {
    if (i % B == 0) {
      for (int j = 0; j <= (n - 1) / B; ++j) {
        lazy_history[i / B][j] = lazy[j];
        maxi_history[i / B][j] = maxi[j];
      }
    }
    int r = order[i], l = max(0, r - m + 1);
    if (l / B == r / B) {
      partial[i].push_back({l, r, 0});
    } else {
      if (l % B != 0) {
        int a = l - l % B + B;
        partial[i].push_back({l, a - 1, 0});
        l = a;
      }
      if (r % B != B - 1) {
        int b = r - r % B - 1;
        partial[i].push_back({b + 1, r, 0});
        r = b;
      }
      whole[i].push_back({l / B, r / B});
    }
    for (auto &[a, b, x] : partial[i]) {
      for (int j = a; j <= b; ++j) {
        maxi[j / B] = max(maxi[j / B], ++quality[j]);
      }
      x = maxi[a / B];
      updates[a / B].push_back({a % B, b % B, i});
    }
    for (auto [a, b] : whole[i]) {
      for (int j = a; j <= b; ++j) {
        ++lazy[j];
      }
    }
  }
  int s, ans = 0;
  cin >> s;
  while (s--) {
    int l, r, q;
    cin >> l >> r >> q;
    --l, --r;
    q = lower_bound(order, order + n, q ^ ans,
                    [&](int a, int x) { return x <= song[a]; }) -
        order;
    fill(lazy, lazy + (n - 1) / B + 1, 0);
    for (int i = 0; i <= (n - 1) / B; ++i) {
      maxi[i] = maxi_history[q / B][i];
      lazy[i] += lazy_history[q / B][i];
      lazy[i + 1] -= lazy_history[q / B][i];
    }
    for (int i = q - q % B; i < q; ++i) {
      for (auto [a, b, c] : partial[i]) {
        maxi[a / B] = c;
      }
      for (auto [a, b] : whole[i]) {
        ++lazy[a];
        --lazy[b + 1];
      }
    }
    for (int i = 1; i <= (n - 1) / B; ++i) {
      lazy[i] += lazy[i - 1];
    }
    ans = 0;
    vector<array<int, 2>> segments;
    if (l / B == r / B) {
      segments.push_back({l, r});
    } else {
      int a = l - l % B + B, b = r - r % B - 1;
      segments.push_back({l, a - 1});
      segments.push_back({b + 1, r});
      for (int i = a / B; i <= b / B; ++i) {
        ans = max(ans, lazy[i] + maxi[i]);
      }
    }
    for (auto [u, v] : segments) {
      fill(maxi, maxi + B, 0);
      for (auto [a, b, i] : updates[u / B]) {
        if (i < q) {
          ++maxi[a];
          if (b + 1 < B) {
            --maxi[b + 1];
          }
        }
      }
      for (int i = 0; i < B; ++i) {
        if (i > 0) {
          maxi[i] += maxi[i - 1];
        }
        if (i >= u % B && i <= v % B) {
          ans = max(ans, lazy[u / B] + maxi[i]);
        }
      }
    }
    ans = m - ans;
    cout << ans << "\n";
  }
}

```