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;
inline int Read() {
  char c = getchar();
  int num = 0;
  while ('0' > c || c > '9') c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  return (num);
}
int wr[20], wt;
inline void Write(int x) {
  if (!x) {
    puts("0");
    return;
  }
  for (; x; x /= 10) wr[++wt] = x % 10;
  for (; wt; wt--) putchar('0' + wr[wt]);
  puts("");
}
int ak[50010 * 4], n, m, q;
vector<pair<int, int> > ad[50010 * 4];
inline void Ins(int x, int l, int r, int a, int b, pair<int, int> c) {
  int md = (l + r) / 2;
  if (l == a && r == b) {
    ad[x].push_back(c);
    return;
  }
  if (b <= md) Ins(x * 2, l, md, a, b, c);
  if (md < a) Ins(x * 2 + 1, md + 1, r, a, b, c);
  if (a <= md && md < b) {
    Ins(x * 2, l, md, a, md, c);
    Ins(x * 2 + 1, md + 1, r, md + 1, b, c);
  }
}
inline void Cov(int x, int l, int r, int a) {
  int md = (l + r) / 2;
  if (l == r) {
    ak[x] = 1;
    return;
  }
  if (a <= md)
    Cov(x * 2, l, md, a);
  else
    Cov(x * 2 + 1, md + 1, r, a);
}
long long F[20][2010];
inline void Dfs(int x, int l, int r, int d) {
  for (int i = 0; i <= m; i++) F[d][i] = F[d - 1][i];
  for (int i = 0; i < ad[x].size(); i++) {
    pair<int, int> ii = ad[x][i];
    for (int j = m; j >= ii.second; j--)
      F[d][j] = max(F[d][j], F[d][j - ii.second] + ii.first);
  }
  if (l == r) {
    if (ak[x]) {
      int as = 0;
      for (int w = 1, i = 1; i <= m; i++, w = w * 1ll * 10000019 % 1000000007)
        as = (as + w * 1ll * F[d][i] % 1000000007) % 1000000007;
      Write(as);
    }
    return;
  }
  int md = (l + r) / 2;
  Dfs(x * 2, l, md, d + 1);
  Dfs(x * 2 + 1, md + 1, r, d + 1);
}
pair<int, int> a[50010];
int st[50010];
int main() {
  n = Read(), m = Read();
  for (int i = 1; i <= n; i++)
    a[i].first = Read(), a[i].second = Read(), st[i] = 1;
  q = Read();
  for (int i = 1; i <= q; i++) {
    int op = Read();
    if (op == 1) st[++n] = i, a[n].first = Read(), a[n].second = Read();
    if (op == 2) {
      int x = Read();
      Ins(1, 1, q, st[x], i, a[x]), st[x] = 0;
    }
    if (op == 3) Cov(1, 1, q, i);
  }
  for (int i = 1; i <= n; i++)
    if (st[i]) Ins(1, 1, q, st[i], q, a[i]);
  Dfs(1, 1, q, 1);
}
