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 int NMAX = 15005, QMAX = 35005, KMAX = 1005;
const int P = 1e7 + 19, Q = 1e9 + 7;
int value[NMAX], weight[NMAX];
int opElem[QMAX], opType[QMAX];
int qpos[QMAX], answer[QMAX];
int64_t rdp[KMAX], adp[KMAX];
void insert(vector<int>& a, const multiset<int>& b) {
  a.insert(a.end(), b.begin(), b.end());
}
void solve(int k, int left, int right, const multiset<int>& add) {
  int64_t* sdp = new int64_t[NMAX];
  memcpy(sdp, rdp, sizeof rdp);
  for (int p : add) {
    memcpy(adp, rdp, sizeof adp);
    for (int i = 0; i < weight[p]; ++i) rdp[i] = adp[i];
    for (int i = weight[p]; i <= k; ++i)
      rdp[i] = max(adp[i], adp[i - weight[p]] + value[p]);
  }
  if (left == right) {
    int ans = 0;
    for (int i = k; i > 0; --i) {
      ans = ((int64_t)ans * P + rdp[i]) % Q;
    }
    answer[left] = ans;
  } else {
    int mid = (left + right) / 2;
    {
      multiset<int> nadd;
      for (int i = qpos[mid]; i < qpos[right]; ++i) {
        if (opType[i] == 1) {
          nadd.insert(opElem[i]);
        }
      }
      for (int i = qpos[left]; i < qpos[right]; ++i) {
        if (opType[i] == 0) {
          auto it = nadd.find(opElem[i]);
          if (it != nadd.end()) nadd.erase(it);
        }
      }
      solve(k, left, mid, nadd);
    }
    {
      multiset<int> nadd;
      for (int i = qpos[left]; i < qpos[mid + 1]; ++i) {
        if (opType[i] == 0) {
          nadd.insert(opElem[i]);
        } else {
          auto it = nadd.find(opElem[i]);
          if (it != nadd.end()) nadd.erase(it);
        }
      }
      for (int i = qpos[mid + 1]; i < qpos[right]; ++i) {
        if (opType[i] == 1) {
          auto it = nadd.find(opElem[i]);
          if (it != nadd.end()) nadd.erase(it);
        }
      }
      solve(k, mid + 1, right, nadd);
    }
  }
  memcpy(rdp, sdp, sizeof rdp);
  delete[] sdp;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  int curr_op = 0;
  for (int i = 0; i < n; ++i) {
    cin >> value[i] >> weight[i];
    opElem[curr_op] = i;
    opType[curr_op] = 0;
    curr_op++;
  }
  int q;
  cin >> q;
  int cntq = 0;
  for (int i = 0; i < q; ++i) {
    int type;
    cin >> type;
    switch (type) {
      case 1:
        cin >> value[n] >> weight[n];
        opElem[curr_op] = n++;
        opType[curr_op] = 0;
        curr_op++;
        break;
      case 2:
        cin >> opElem[curr_op];
        opElem[curr_op]--;
        opType[curr_op] = 1;
        curr_op++;
        break;
      case 3:
        qpos[cntq++] = curr_op;
        break;
      default:
        throw 13;
    }
  }
  multiset<int> add;
  for (int i = 0; i < qpos[0]; ++i) {
    if (opType[i] == 0) {
      add.insert(opElem[i]);
    } else {
      add.erase(add.find(opElem[i]));
    }
  }
  for (int i = qpos[0]; i < qpos[cntq - 1]; ++i) {
    if (opType[i] == 1) {
      auto it = add.find(opElem[i]);
      if (it != add.end()) add.erase(it);
    }
  }
  solve(k, 0, cntq - 1, add);
  for (int i = 0; i < cntq; ++i) cout << answer[i] << '\n';
}
