Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, K, Q;
int itemValue[15000];
int itemMass[15000];
int itemBegin[15000];
int itemEnd[15000];
int cnt[15000];
int queryTime[32768];
int s[32768][1001];
int numAdded[32768];
int numNeeded[32768];
int etp[100000];
int ei[100000];
void add(int query, int item) {
  for (int c = K; c >= itemMass[item]; c--)
    s[query][c] =
        max(s[query][c], s[query][c - itemMass[item]] + itemValue[item]);
}
void copy(int query1, int query2) {
  for (int c = K; c >= 0; c--) s[query1][c] = s[query2][c];
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> K;
  int T = 0;
  int numItems = 0;
  int numQueries = 0;
  for (; T < N; T++) {
    cin >> itemValue[numItems] >> itemMass[numItems];
    itemBegin[numItems] = T, itemEnd[numItems] = 100000;
    etp[T] = 0;
    ei[T] = numItems;
    numItems++;
  }
  cin >> Q;
  int tp, x;
  int curItems = N;
  for (; T < N + Q; T++) {
    cin >> tp;
    if (tp == 1) {
      cin >> itemValue[numItems] >> itemMass[numItems];
      itemBegin[numItems] = T, itemEnd[numItems] = 100000;
      etp[T] = 0;
      ei[T] = numItems;
      numItems++;
      curItems++;
    } else if (tp == 2) {
      cin >> x;
      itemEnd[x - 1] = T;
      etp[T] = 0;
      ei[T] = x - 1;
      curItems--;
    } else {
      queryTime[numQueries] = T;
      etp[T] = 1;
      ei[T] = numQueries;
      numNeeded[numQueries] = curItems;
      numQueries++;
    }
  }
  int onum = numQueries;
  while (numQueries != (numQueries & -numQueries)) {
    etp[T] = 1;
    ei[T] = numQueries;
    queryTime[numQueries] = T;
    numNeeded[numQueries] = curItems;
    numQueries++, T++;
  }
  assert(T < 100000);
  assert(numQueries <= 32768);
  assert(numItems <= 15000);
  for (int i = 0; i < queryTime[0]; i++) {
    if (itemEnd[ei[i]] > queryTime[numQueries - 1]) {
      add(0, ei[i]);
      numAdded[0]++;
    }
  }
  for (int d = numQueries / 2; d >= 1; d /= 2) {
    for (int i = 0; i < numItems; i++) cnt[i] = 0;
    int left = 0;
    for (int i = numQueries - d; i >= 0; i -= d, left ^= 1) {
      if (left) {
        for (int j = queryTime[i + d - 1] + 1; j < queryTime[i + 2 * d - 1];
             j++)
          if (etp[j] == 0) {
            if (itemBegin[ei[j]] < queryTime[i]) {
              add(i, ei[j]);
              cnt[ei[j]]++;
              numAdded[i]++;
            }
          }
      } else {
        copy(i, i - d);
        numAdded[i] = numAdded[i - d];
        for (int j = queryTime[i - d] + 1; j < queryTime[i]; j++)
          if (etp[j] == 0) {
            if (itemEnd[ei[j]] > queryTime[i + d - 1]) {
              add(i, ei[j]);
              cnt[ei[j]]++;
              numAdded[i]++;
            }
          }
      }
    }
  }
  for (int i = 0; i < T; i++) {
    if (etp[i] == 1 && ei[i] < onum) {
      long long ans = 0;
      for (int c = K; c >= 1; c--)
        ans = (ans * 10000019 + s[ei[i]][c]) % 1000000007;
      cout << ans << '\n';
    }
  }
}