There's a famous museum in the city where Kleofáš lives. In the museum, n exhibits (numbered 1 through n) had been displayed for a long time; the i-th of those exhibits has value vi and mass wi. 

Then, the museum was bought by a large financial group and started to vary the exhibits. At about the same time, Kleofáš... gained interest in the museum, so to say.

You should process q events of three types:

  * type 1 — the museum displays an exhibit with value v and mass w; the exhibit displayed in the i-th event of this type is numbered n + i (see sample explanation for more details)
  * type 2 — the museum removes the exhibit with number x and stores it safely in its vault
  * type 3 — Kleofáš visits the museum and wonders (for no important reason at all, of course): if there was a robbery and exhibits with total mass at most m were stolen, what would their maximum possible total value be?



For each event of type 3, let s(m) be the maximum possible total value of stolen exhibits with total mass  ≤ m.

Formally, let D be the set of numbers of all exhibits that are currently displayed (so initially D = {1, ..., n}). Let P(D) be the set of all subsets of D and let 

<image>

Then, s(m) is defined as 

<image>

Compute s(m) for each <image>. Note that the output follows a special format.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 1000) — the initial number of exhibits in the museum and the maximum interesting mass of stolen exhibits. 

Then, n lines follow. The i-th of them contains two space-separated positive integers vi and wi (1 ≤ vi ≤ 1 000 000, 1 ≤ wi ≤ 1000) — the value and mass of the i-th exhibit.

The next line contains a single integer q (1 ≤ q ≤ 30 000) — the number of events.

Each of the next q lines contains the description of one event in the following format:

  * 1 v w — an event of type 1, a new exhibit with value v and mass w has been added (1 ≤ v ≤ 1 000 000, 1 ≤ w ≤ 1000)
  * 2 x — an event of type 2, the exhibit with number x has been removed; it's guaranteed that the removed exhibit had been displayed at that time
  * 3 — an event of type 3, Kleofáš visits the museum and asks his question



There will be at most 10 000 events of type 1 and at least one event of type 3.

Output

As the number of values s(m) can get large, output the answers to events of type 3 in a special format.

For each event of type 3, consider the values s(m) computed for the question that Kleofáš asked in this event; print one line containing a single number 

<image>

where p = 107 + 19 and q = 109 + 7.

Print the answers to events of type 3 in the order in which they appear in the input.

Examples

Input

3 10
30 4
60 6
5 1
9
3
1 42 5
1 20 3
3
2 2
2 4
3
1 40 6
3


Output

556674384
168191145
947033915
181541912


Input

3 1000
100 42
100 47
400 15
4
2 2
2 1
2 3
3


Output

0

Note

In the first sample, the numbers of displayed exhibits and values s(1), ..., s(10) for individual events of type 3 are, in order: 

<image> <image> <image> <image>

The values of individual exhibits are v1 = 30, v2 = 60, v3 = 5, v4 = 42, v5 = 20, v6 = 40 and their masses are w1 = 4, w2 = 6, w3 = 1, w4 = 5, w5 = 3, w6 = 6.

In the second sample, the only question is asked after removing all exhibits, so s(m) = 0 for any m.
#include <bits/stdc++.h>
using namespace std;
const long long P = 10 * 1000 * 1000 + 19, Q = 1000 * 1000 * 1000 + 7;
int k;
vector<int> best;
void update(int w, int val) {
  for (int i = k; i >= w; i--) {
    best[i] = max(best[i], best[i - w] + val);
  }
}
int calc(int x) {
  long long ppow = 1, ans = 0;
  for (int i = 1; i <= x; i++) {
    ans = (ans + 1LL * best[i] * ppow) % Q;
    ppow = (ppow * P) % Q;
  }
  return (int)(ans % Q);
}
struct query {
  int type;
  int id;
  int val;
  int l, r;
  query(int _type, int _id, int _l, int _r, int _val = 0) {
    type = _type;
    id = _id;
    l = _l;
    r = _r;
    val = _val;
  }
};
int w[100007], val[100007], open[100007], close[100007];
int answers[100007];
bool inside(int l1, int r1, int l2, int r2) {
  if (l1 >= l2 && l1 <= r2) return true;
  if (r1 >= l2 && r1 <= r2) return true;
  return false;
}
void go(vector<query>& q, int l, int r) {
  vector<int> tmp = best;
  for (int i = 0; i < q.size(); i++) {
    if (q[i].type == 1 && q[i].l <= l && q[i].r >= r) {
      update(w[q[i].id], val[q[i].id]);
    }
  }
  if (l == r) {
    for (int i = 0; i < q.size(); i++) {
      if (q[i].type == 2 && q[i].l == l && q[i].r == r) {
        answers[q[i].id] = calc(q[i].val);
      }
    }
    best = tmp;
    return;
  }
  vector<query> q2;
  for (int i = 0; i < q.size(); i++) {
    if (q[i].type == 1 && q[i].l <= l && q[i].r >= r) {
      ;
      ;
    } else if ((q[i].type == 2 && q[i].l <= l && q[i].r >= r) ||
               (q[i].l <= r && q[i].r >= l)) {
      q2.push_back(q[i]);
    }
  }
  int m = (l + r) / 2;
  go(q2, l, m);
  go(q2, m + 1, r);
  best = tmp;
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  vector<query> q;
  int timer = 0;
  for (int i = 0; i <= k; i++) {
    best.push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    timer++;
    scanf("%d%d", &val[i], &w[i]);
    open[i] = 1;
    q.push_back(query(1, i, timer, -1));
  }
  int cntId = n, cntQ = 0;
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    timer++;
    int t;
    scanf("%d", &t);
    if (t == 1) {
      cntId++;
      scanf("%d%d", &val[cntId], &w[cntId]);
      open[cntId] = timer;
      q.push_back(query(1, cntId, open[cntId], -1));
    } else if (t == 2) {
      int id;
      scanf("%d", &id);
      close[id] = timer;
    } else if (t == 3) {
      cntQ++;
      q.push_back(query(2, cntQ, timer, timer, k));
    }
  }
  for (int i = 1; i <= cntId; i++) {
    if (close[i] == 0) {
      timer++;
      close[i] = timer;
    }
  }
  for (int i = 0; i < q.size(); i++) {
    if (q[i].type == 1) {
      q[i].l = open[q[i].id];
      q[i].r = close[q[i].id];
    }
  }
  go(q, 1, timer);
  for (int i = 1; i <= cntQ; i++) {
    printf("%d\n", answers[i]);
  }
}
