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;
const int OFF = (1 << 15);
int k, n, Q, E;
int beg[15010], fin[15010], val[15010], w[15010];
vector<int> tour[2 * OFF + 10];
int sol[OFF + 10][1010];
int OUT[OFF];
long long p = 1e7 + 19, mod = 1e9 + 7;
void solve(int x) {
  if (x >= OFF) {
    for (int i = (0); i < (k + 1); i++) sol[OFF][i] = sol[x / 2][i];
    for (int i = (0); i < (tour[x].size()); i++) {
      int tr = tour[x][i];
      for (int j = (k); j >= (0); j--) {
        if (j >= w[tr])
          sol[OFF][j] = max(sol[OFF][j], sol[OFF][j - w[tr]] + val[tr]);
      }
    }
    long long out = 0, um = 1;
    for (int j = (1); j < (k + 1); j++) {
      out = (out + ((um * (long long)sol[OFF][j]) % mod)) % mod;
      um = (um * p) % mod;
    }
    OUT[x - OFF] = out;
    return;
  }
  for (int i = (0); i < (k + 1); i++) sol[x][i] = sol[x / 2][i];
  for (int i = (0); i < (tour[x].size()); i++) {
    int tr = tour[x][i];
    for (int j = (k); j >= (0); j--) {
      if (j >= w[tr]) sol[x][j] = max(sol[x][j], sol[x][j - w[tr]] + val[tr]);
    }
  }
  solve(x * 2);
  solve(x * 2 + 1);
  return;
}
void add(int a, int b, int num, int lo, int hi, int x) {
  if (a <= lo && hi <= b) {
    tour[x].push_back(num);
    return;
  }
  if (a >= hi || b <= lo) return;
  int mi = (lo + hi) / 2;
  add(a, b, num, lo, mi, x * 2);
  add(a, b, num, mi, hi, x * 2 + 1);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(fin, -1, sizeof fin);
  cin >> n >> k;
  E = n;
  for (int i = (0); i < (n); i++) {
    cin >> val[i] >> w[i];
    beg[i] = 0;
  }
  int q;
  cin >> q;
  for (int i = (0); i < (q); i++) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      cin >> val[E] >> w[E];
      beg[E] = Q;
      E++;
    }
    if (ty == 2) {
      int x;
      cin >> x;
      x--;
      fin[x] = Q;
    }
    if (ty == 3) {
      Q++;
    }
  }
  for (int i = (0); i < (E); i++) {
    if (fin[i] == -1) fin[i] = Q;
    add(beg[i], fin[i], i, 0, OFF, 1);
  }
  solve(1);
  for (int i = (0); i < (Q); i++) {
    cout << OUT[i] << "\n";
  }
  return 0;
}