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;
long long int modulo = 1e9 + 7;
long long int base = 1e7 + 19;
const int limite = 1 << 15;
int k;
void anyade(int maximo[1001], int v, int w) {
  for (int i = k; i >= w; i--) maximo[i] = max(maximo[i], maximo[i - w] + v);
}
void copia(int maximo[1001], int nextmaximo[1001]) {
  for (int i = 0; i <= k; i++) nextmaximo[i] = maximo[i];
}
int calcula(int maximo[1001]) {
  int suma = 0;
  long long int prod = 1;
  for (int i = 1; i <= k; i++) {
    suma = (suma + prod * maximo[i]) % modulo;
    prod = prod * base % modulo;
  }
  return suma;
}
int n;
int maximo[limite][1001];
int v[limite];
int w[limite];
int le[limite];
int ri[limite];
int tipo[limite];
int elem[limite];
int entra[limite];
int sale[limite];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
    entra[i] = -1;
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> tipo[i];
    if (tipo[i] == 1) {
      n++;
      elem[i] = n;
      entra[elem[i]] = i;
      cin >> v[elem[i]] >> w[elem[i]];
    } else if (tipo[i] == 2) {
      cin >> elem[i];
      sale[elem[i]] = i;
    }
  }
  for (int i = 1; i <= n; i++)
    if (sale[i] == 0) sale[i] = limite;
  for (int i = 1; i <= n; i++)
    if (entra[i] == -1 and sale[i] == limite) anyade(maximo[0], v[i], w[i]);
  le[0] = 0;
  ri[0] = limite;
  int amount = 1;
  do {
    amount *= 2;
    for (int i = amount - 1; i >= 0; i--) {
      int p = i >> 1;
      copia(maximo[p], maximo[i]);
      if (i & 1) {
        ri[i] = ri[p];
        le[i] = (le[p] + ri[p]) / 2;
        for (int j = le[p]; j < le[i]; j++) {
          if (tipo[j] == 1 and sale[elem[j]] >= ri[i]) {
            anyade(maximo[i], v[elem[j]], w[elem[j]]);
          }
        }
      } else {
        le[i] = le[p];
        int rip = ri[p];
        ri[i] = (le[p] + ri[p]) / 2;
        for (int j = ri[i]; j < rip; j++) {
          if (tipo[j] == 2 and entra[elem[j]] < le[i]) {
            anyade(maximo[i], v[elem[j]], w[elem[j]]);
          }
        }
      }
    }
  } while (amount < limite);
  for (int i = 1; i <= q; i++)
    if (tipo[i] == 3) cout << calcula(maximo[i]) << endl;
}
