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;
int n, k, q;
int cnt;
const int N = 50000;
const int P = 20;
int vi[N], wi[N], ti[N], tj[N];
int lb[N], rb[N], lc[N], rc[N], tot;
vector<int> V[N];
int build(int l, int r) {
  int t = ++tot;
  lb[t] = l;
  rb[t] = r;
  if (l < r) {
    lc[t] = build(l, (l + r) / 2);
    rc[t] = build((l + r) / 2 + 1, r);
  }
  return t;
}
void modify(int t, int l, int r, int k) {
  if (l > r) return;
  if (l <= lb[t] && rb[t] <= r) {
    V[t].push_back(k);
    return;
  }
  if (l <= rb[lc[t]]) modify(lc[t], l, r, k);
  if (r >= lb[rc[t]]) modify(rc[t], l, r, k);
}
int top;
int F[P][N];
void dfs(int t) {
  top++;
  for (int i = 1; i <= k; ++i) F[top][i] = F[top - 1][i];
  for (int i = 0; i < V[t].size(); ++i)
    for (int j = k - wi[V[t][i]]; j >= 0; --j)
      F[top][j + wi[V[t][i]]] =
          max(F[top][j + wi[V[t][i]]], F[top][j] + vi[V[t][i]]);
  if (lb[t] == rb[t]) {
    int p = 1e7 + 19, q = 1e9 + 7;
    int tmp1 = 1, res = 0;
    for (int i = 1; i <= k; ++i) {
      res = (1ll * tmp1 * F[top][i] + res) % q;
      tmp1 = 1ll * tmp1 * p % q;
    }
    cout << res << endl;
  } else {
    dfs(lc[t]);
    dfs(rc[t]);
  }
  top--;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> vi[i] >> wi[i], ti[i] = 1, tj[i] = -1;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int opt, p;
    cin >> opt;
    if (opt == 1)
      n++, cin >> vi[n] >> wi[n], ti[n] = cnt + 1, tj[n] = -1;
    else if (opt == 2)
      cin >> p, tj[p] = cnt;
    else
      cnt++;
  }
  for (int i = 1; i <= n; ++i)
    if (tj[i] == -1) tj[i] = cnt;
  build(1, cnt);
  for (int i = 1; i <= n; ++i) modify(1, ti[i], tj[i], i);
  dfs(1);
}
