601_E. A Museum Robbery

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.

Example 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

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


Example output:
                                                       556674384
                                                       168191145
                                                       947033915
                                                       181541912

                                                               0


Tags: data structures
dp
Difficulty rating: 2800

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, k, qn;
int v[15300], w[15300];
int st[15300], ed[15300];
long long P = 1e7 + 19, mod = 1e9 + 7;
void dfs(int l, int r, bitset<15100> now, vector<int> f) {
  for (int i = 1; i <= n; i++)
    if (now[i] == 0 && st[i] <= l && ed[i] >= r) {
      now[i] = 1;
      for (int j = k - w[i]; j >= 0; j--)
        f[j + w[i]] = max(f[j + w[i]], f[j] + v[i]);
    }
  if (l == r) {
    long long ans = 0;
    long long p = 1;
    for (int i = 1; i <= k; i++) {
      ans = (ans + f[i] * p) % mod;
      p = p * P % mod;
    }
    cout << ans << endl;
  } else {
    int mid = (l + r) / 2;
    dfs(l, mid, now, f);
    dfs(mid + 1, r, now, f);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i], &w[i]);
    st[i] = 0;
    ed[i] = 100000;
  }
  int m;
  cin >> m;
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      n++;
      st[n] = qn + 1;
      ed[n] = 1000000;
      scanf("%d%d", &v[n], &w[n]);
    } else if (type == 2) {
      int x;
      scanf("%d", &x);
      ed[x] = qn;
    } else
      qn++;
  }
  dfs(1, qn, bitset<15100>(), vector<int>(k + 1, 0));
}