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;
const int p = 1e7 + 19, q = 1e9 + 7;
int dp[1 << 15][1111], ans;
int n, k, l[22222], r[22222], cnt, v[22222], w[22222], tim, Q, tp, x, num,
    t[33333], sz;
bool f;
vector<pair<int, int> > tree[1 << 15];
void update(int beg, int en, int l, int r, int node, int i) {
  if (l > en || r < beg) return;
  if (l >= beg && r <= en) {
    tree[node].push_back(make_pair(v[i], w[i]));
    return;
  }
  int mid = (l + r) >> 1;
  update(beg, en, l, mid, node << 1, i);
  update(beg, en, mid + 1, r, (node << 1) | 1, i);
}
void dfs(int l, int r, int node) {
  for (int i = 1; i <= k; i++) dp[node][i] = dp[node >> 1][i];
  for (int i = 0; i < tree[node].size(); i++) {
    int v = tree[node][i].first, w = tree[node][i].second;
    for (int j = k; j >= w; j--)
      dp[node][j] = max(dp[node][j], dp[node][j - w] + v);
  }
  if (l == r) return;
  int mid = (l + r) >> 1;
  dfs(l, mid, node << 1);
  dfs(mid + 1, r, (node << 1) | 1);
}
int main() {
  scanf("%d%d", &n, &k);
  sz = 1 << 14;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i], &w[i]);
    l[i] = 1;
  }
  cnt = n;
  tim = 1;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &tp);
    if (tp == 1) {
      cnt++;
      scanf("%d%d", &v[cnt], &w[cnt]);
      l[cnt] = tim + 1;
      f = 0;
    } else if (tp == 2) {
      scanf("%d", &x);
      r[x] = tim;
      f = 0;
    } else {
      ++num;
      if (f)
        t[num] = tim;
      else
        t[num] = ++tim;
      f = 1;
    }
  }
  for (int i = 1; i <= cnt; i++) {
    if (!r[i]) r[i] = tim;
    update(l[i], r[i], 1, sz, 1, i);
  }
  dfs(1, sz, 1);
  for (int i = 1; i <= num; i++) {
    ans = 0;
    for (int j = k; j >= 1; j--) {
      ans = (1ll * ans * p + dp[sz - 1 + t[i]][j]) % q;
    }
    printf("%d\n", ans);
  }
  return 0;
}
