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;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
int k, cnt, v[15000 + 5], w[15000 + 5], u[15000 + 5], Q[32768 + 5],
    f[20][32768 * 2 + 5];
vector<int> t[32768 * 2 + 5];
void ins(int l, int r, int x) {
  for (l += 32768 - 1, r += 32768 + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) t[l + 1].push_back(x);
    if (r & 1) t[r - 1].push_back(x);
  }
}
void solve(int x, int dep) {
  for (int i = 0; i <= k; ++i) f[dep][i] = f[dep - 1][i];
  for (int i = 0; i < t[x].size(); ++i)
    for (int j = k; j >= w[t[x][i]]; --j)
      f[dep][j] = max(f[dep][j], f[dep][j - w[t[x][i]]] + v[t[x][i]]);
  if (x < 32768)
    solve(x << 1, dep + 1), solve(x << 1 | 1, dep + 1);
  else if (Q[x - 32768] > 2) {
    int ans = 0, i;
    for (i = k; i; --i) ans = (1LL * ans * 10000019 + f[dep][i]) % 1000000007;
    printf("%d\n", ans);
  }
}
int main() {
  int n, q, i;
  cnt = n = read();
  k = read();
  for (i = 1; i <= n; ++i) v[i] = read(), w[i] = read(), u[i] = 1;
  for (q = read(), i = 1; i <= q; ++i) {
    Q[i] = read();
    if (Q[i] == 1) v[++cnt] = read(), w[cnt] = read(), u[cnt] = i;
    if (Q[i] == 2) n = read(), ins(u[n], i, n), u[n] = 0;
  }
  for (i = 1; i <= cnt; ++i)
    if (u[i]) ins(u[i], q, i);
  solve(1, 1);
}