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 off = 1 << 15, mod = 1e9 + 7, P = 1e7 + 19, MAX = 15005, MAXQ = 35005;
int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
int mul(int x, int y) { return (long long)x * y % mod; }
pair<int, int> A[MAX];
vector<pair<int, int> > T[2 * off];
int n, k;
void update(int x, int lo, int hi, int a, int b, int i) {
  if (lo >= a && hi <= b) {
    T[x].push_back(A[i]);
    return;
  }
  if (lo >= b || hi <= a) {
    return;
  }
  int mid = (lo + hi) >> 1;
  update(x * 2 + 0, lo, mid, a, b, i);
  update(x * 2 + 1, mid, hi, a, b, i);
}
int cookie;
int lo[MAX], hi[MAX];
stack<vector<int> > knapsack;
int calc() {
  int ret = 0;
  int p = 1;
  for (int i = 1; i < k + 1; ++i) {
    ret = add(ret, mul(knapsack.top()[i], p));
    p = mul(p, P);
  }
  return ret;
}
int sol[MAXQ];
void dfs(int x, int lo, int hi) {
  if (lo >= cookie) return;
  vector<int> new_knap, before;
  new_knap = knapsack.top();
  for (auto p : T[x]) {
    before = new_knap;
    for (int i = 0; i < k + 1; ++i) {
      new_knap[i] = before[i];
      if (i - p.second >= 0) {
        new_knap[i] = max(new_knap[i], add(before[i - p.second], p.first));
      }
    }
  }
  knapsack.push(new_knap);
  if (lo == hi - 1) {
    sol[lo] = calc();
    knapsack.pop();
    return;
  }
  int mid = (lo + hi) >> 1;
  dfs(x * 2 + 0, lo, mid);
  dfs(x * 2 + 1, mid, hi);
  knapsack.pop();
}
int main() {
  fill(hi, hi + MAX, off);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> A[i].first >> A[i].second;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int type;
    cin >> type;
    if (type == 1) {
      cin >> A[n].first >> A[n].second;
      lo[n] = cookie;
      n++;
    }
    if (type == 2) {
      int x;
      cin >> x;
      hi[x - 1] = cookie;
    }
    if (type == 3) {
      cookie++;
    }
  }
  for (int i = 0; i < n; ++i) {
    update(1, 0, off, lo[i], hi[i], i);
  }
  vector<int> v;
  v.resize(k + 1, 0);
  knapsack.push(v);
  dfs(1, 0, off);
  for (int i = 0; i < cookie; ++i) {
    cout << sol[i] << endl;
  }
}
