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 maxn = 15005, maxq = 30005, maxk = 1005;
vector<pair<int, int> > lk[maxq * 4];
pair<int, int> info[maxn];
int qr[maxq], sav[20][maxk];
int ans[maxq][maxk], st[maxn], en[maxn], n, k;
void push(int l, int r, int i, int j, int c, int jd) {
  if (j < l || i > r || qr[r] == qr[l - 1]) return;
  if (i <= l && r <= j)
    lk[jd].push_back(info[c]);
  else {
    int mid = l + r >> 1;
    push(l, mid, i, j, c, jd << 1), push(mid + 1, r, i, j, c, jd << 1 | 1);
  }
}
void build(int dep, int l, int r, int jd) {
  if (qr[r] == qr[l - 1]) return;
  for (int i = 0; i <= k; i++) sav[dep][i] = sav[dep - 1][i];
  for (int i = 1; i <= k; i++) sav[dep][i] = max(sav[dep][i], sav[dep][i - 1]);
  for (auto p : lk[jd])
    for (int i = k; i >= p.second; i--)
      sav[dep][i] = max(sav[dep][i], sav[dep][i - p.second] + p.first);
  if (l == r) {
    for (int i = 0; i <= k; i++) ans[l][i] = max(ans[l][i], sav[dep][i]);
    return;
  }
  int mid = l + r >> 1;
  build(dep + 1, l, mid, jd << 1);
  build(dep + 1, mid + 1, r, jd << 1 | 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++)
    scanf("%d%d", &info[i].first, &info[i].second), st[i] = 1;
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int ty, v, w;
    scanf("%d", &ty);
    if (ty == 1) {
      ++n;
      scanf("%d%d", &info[n].first, &info[n].second);
      st[n] = i;
    }
    if (ty == 2) {
      scanf("%d", &v);
      en[v] = i;
    }
    if (ty == 3) qr[i] = 1;
    qr[i] += qr[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    if (!en[i]) en[i] = q;
    push(1, q, st[i], en[i], i, 1);
  }
  build(1, 1, q, 1);
  for (int i = 1; i <= q; i++)
    if (qr[i] != qr[i - 1]) {
      int p = int(1e7) + 19, q = int(1e9) + 7;
      int v = 0;
      for (int m = k; m; m--) v = (v * 1ll * p + ans[i][m]) % q;
      printf("%d\n", v);
    }
  return 0;
}
