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 N = 30005, M = 1e9 + 7, P = 1e7 + 19;
int n, m, a[N], b[N], last[N], first[N], opt, flag[N], tot, Stack[20][N], x, q;
struct Tree {
  int l, r;
  vector<pair<int, int> > v;
} T[N * 4];
void build(int x, int l, int r) {
  T[x].l = l;
  T[x].r = r;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
}
void insert(int x, int l, int r, pair<int, int> z) {
  if (T[x].l > r || l > T[x].r) return;
  if (T[x].l >= l && T[x].r <= r) {
    T[x].v.push_back(z);
    return;
  }
  insert(x * 2, l, r, z);
  insert(x * 2 + 1, l, r, z);
}
inline void upd(int &x, int y) {
  if (x < y) x = y;
}
void solve(int x) {
  tot++;
  memcpy(Stack[tot], Stack[tot - 1], sizeof Stack[tot]);
  for (pair<int, int> i : T[x].v)
    for (int j = m; j >= i.second; j--)
      upd(Stack[tot][j], Stack[tot][j - i.second] + i.first);
  if (T[x].l == T[x].r) {
    if (flag[T[x].l]) {
      int ans = 0, s = 1;
      for (int i = 1; i <= m; i++)
        ans = (ans + (long long)s * Stack[tot][i]) % M,
        s = (long long)s * P % M;
      printf("%d\n", ans);
    }
  } else {
    solve(x * 2);
    solve(x * 2 + 1);
  }
  tot--;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]), first[i] = 1;
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) last[i] = q;
  build(1, 1, q);
  for (int i = 1; i <= q; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      n++;
      scanf("%d%d", &a[n], &b[n]);
      first[n] = i;
      last[n] = q;
    }
    if (opt == 2) {
      scanf("%d", &x);
      last[x] = i - 1;
    }
    if (opt == 3) flag[i] = 1;
  }
  for (int i = 1; i <= n; i++)
    insert(1, first[i], last[i], make_pair(a[i], b[i]));
  solve(1);
  return 0;
}