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;
struct dp {
  int v[1005];
  dp() {
    for (int i = (0); i < (k + 1); i++) v[i] = 0;
  }
  void add(pair<int, int> l) {
    int x = l.first, w = l.second;
    for (int i = k; i >= w; i--) v[i] = max(v[i], v[i - w] + x);
  }
};
int del[50005];
pair<int, int> ins[50005];
pair<int, int> pt[50005];
int t[50005];
int ind[50005];
int r[50005];
struct rt {
  vector<pair<int, int> > v[50000 * 4];
  void u(int i, int rs, int re, int qs, int qe, pair<int, int> p) {
    if (rs == qs && re == qe)
      v[i].push_back(p);
    else {
      int md = (rs + re) / 2;
      if (qs < md) u(2 * i + 1, rs, md, qs, min(qe, md), p);
      if (qe > md) u(2 * i + 2, md, re, max(qs, md), qe, p);
    }
  }
  void g(int i, int rs, int re, dp f) {
    for (auto p : v[i]) f.add(p);
    if (rs + 1 == re) {
      long long ans = 0;
      long long cp = 1;
      for (int j = (1); j < (k + 1); j++) {
        ans += cp * f.v[j];
        cp = (cp * 10000019ll) % 1000000007ll;
        ans %= 1000000007ll;
      }
      r[rs] = ans;
    } else {
      int md = (rs + re) / 2;
      g(2 * i + 1, rs, md, f);
      g(2 * i + 2, md, re, f);
    }
  }
};
rt a;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n); i++) {
    scanf("%d%d", &ins[i].first, &ins[i].second);
  }
  int q;
  scanf("%d", &q);
  for (int i = (0); i < (50005); i++) del[i] = q;
  int cind = n;
  for (int i = (0); i < (q); i++) {
    scanf("%d", &t[i]);
    if (t[i] == 1) {
      scanf("%d%d", &pt[i].first, &pt[i].second);
      ind[i] = ++cind;
    } else if (t[i] == 2) {
      int x;
      scanf("%d", &x);
      del[x] = i + 1;
    }
  }
  for (int i = (0); i < (n); i++) {
    a.u(0, 0, q, 0, del[i + 1], ins[i]);
  }
  for (int i = (0); i < (q); i++)
    if (t[i] == 1) {
      a.u(0, 0, q, i, del[ind[i]], pt[i]);
    }
  a.g(0, 0, q, dp());
  for (int i = (0); i < (q); i++)
    if (t[i] == 3) printf("%d\n", r[i]);
}
