A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.

Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.

As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.

Input

The first line contains five integers n, k, a, b, and q (1 ≤ k ≤ n ≤ 200 000, 1 ≤ b < a ≤ 10 000, 1 ≤ q ≤ 200 000) — the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.

The next q lines contain the descriptions of the queries. Each query is of one of the following two forms: 

  * 1 di ai (1 ≤ di ≤ n, 1 ≤ ai ≤ 10 000), representing an update of ai orders on day di, or 
  * 2 pi (1 ≤ pi ≤ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi? 



It's guaranteed that the input will contain at least one query of the second type.

Output

For each query of the second type, print a line containing a single integer — the maximum number of orders that the factory can fill over all n days.

Examples

Input

5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3


Output

3
6
4


Input

5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2


Output

7
1

Note

Consider the first sample.

We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.

For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.

For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
#include <bits/stdc++.h>
using namespace std;
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename T>
void read_integer(T &r) {
  bool sign = 0;
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c == '-') {
      sign = 1;
      break;
    }
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') break;
    r = r * 10 + (c - '0');
  }
  if (sign) r = -r;
}
long long binpowmod(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long c = binpowmod(a, b >> 1, mod);
  return (((c * c) % mod) * (b & 1 ? a : 1)) % mod;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long c = binpow(a, b >> 1);
  return c * c * (b & 1 ? a : 1);
}
inline int getbit(int x, int b) { return (x >> b) & 1; }
inline long long getbit(long long x, int b) { return (x >> b) & 1; }
inline unsigned long long getbit(unsigned long long x, int b) {
  return (x >> b) & 1;
}
inline int setbit(int x, int b) { return x | (1 << b); }
inline void _setbit(int &x, int b) { x = setbit(x, b); }
inline long long setbit(long long x, int b) { return x | (1ll << b); }
inline void _setbit(long long &x, int b) { x = setbit(x, b); }
inline int unsetbit(int x, int b) { return x & (INT_MAX - (1 << b)); }
inline void _unsetbit(int &x, int b) { x = unsetbit(x, b); }
inline int countbit(int x) {
  x = x - ((x >> 1) & 0x55555555);
  x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
  return ((x + (x >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline long long countbit(long long x) {
  int p1 = (x >> 32) & ((1ll << 32) - 1);
  int p2 = x & ((1ll << 32) - 1);
  return countbit(p1) + countbit(p2);
}
template <typename T>
void printbit(T x, int len) {
  for (int i = len - 1; i >= 0; i--) printf("%d", getbit(x, i));
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
template <typename A, typename B>
ostream &operator<<(ostream &stream, const pair<A, B> &p) {
  stream << "{" << p.first << "," << p.second << "}";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const vector<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A, typename B>
ostream &operator<<(ostream &stream, const map<A, B> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const set<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const stack<A> &v) {
  stack<A> st;
  stream << "[";
  while (!st.empty()) {
    stream << st.top() << " ";
    st.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const priority_queue<A> &v) {
  priority_queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.top() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
    }
  } while (0);
  return 0;
}
struct SegmentTree {
  vector<long long> tr;
  int n, mx;
  SegmentTree(int _n, int _mx) : n(_n), mx(_mx) { tr.resize(4 * n); }
  void update_impl(int v, int l, int r, int pos, long long d) {
    if (l == r)
      tr[v] = min(1ll * mx, tr[v] + d);
    else {
      int mid = (l + r) >> 1;
      if (pos <= mid)
        update_impl(v + v, l, mid, pos, d);
      else
        update_impl(v + v + 1, mid + 1, r, pos, d);
      tr[v] = tr[v + v] + tr[v + v + 1];
    }
  }
  long long get_sum_impl(int v, int l, int r, int L, int R) {
    if (L > R) return 0;
    if (l == L && r == R)
      return tr[v];
    else {
      int mid = (l + r) >> 1;
      return get_sum_impl(v + v, l, mid, L, min(mid, R)) +
             get_sum_impl(v + v + 1, mid + 1, r, max(mid + 1, L), R);
    }
  }
  long long get_sum(int l, int r) { return get_sum_impl(1, 0, n - 1, l, r); }
  void update(int pos, long long d) { update_impl(1, 0, n - 1, pos, d); }
};
void run() {
  int n, k, a, b, q;
  scanf("%d%d", &n, &k);
  scanf("%d%d%d", &a, &b, &q);
  int t, x, y;
  SegmentTree atree(n + 2, a), btree(n + 2, b);
  while (q--) {
    scanf("%d%d", &t, &x);
    if (t == 1) {
      scanf("%d", &y);
      atree.update(x, y);
      btree.update(x, y);
    } else {
      long long res = btree.get_sum(1, x - 1) + atree.get_sum(x + k, n);
      printf(0 ? "%lld" : "%I64d", res);
      putchar('\n');
    }
  }
}
