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;
using LD = long double;
const int N = 4e4 + 7, inf = 1e9 + 7, MOD = 1e9 + 7;
const long long INF = 2e18 + 7;
const int M = 1 << 16;
int in[N], out[N], val[N], mass[N];
int pot[1007];
int maxMass;
void update(vector<long long>& plecak, int mass, int val) {
  for (int i = plecak.size() - 1; i >= mass; --i)
    plecak[i] = max(plecak[i], plecak[i - mass] + val);
}
void solve(int l, int r, vector<long long> plecak, vector<int> dane) {
  if (r - l == 1) {
    for (auto u : dane)
      if (in[u] <= l && out[u] > r) {
        update(plecak, mass[u], val[u]);
      }
    int res = 0;
    for (int i = 1; i <= maxMass; ++i) {
      res = (res + plecak[i] * pot[i - 1]) % MOD;
    }
    cout << res << '\n';
    return;
  }
  vector<int> nowe;
  for (auto u : dane) {
    if (in[u] <= l && out[u] > r) {
      update(plecak, mass[u], val[u]);
    } else
      nowe.push_back(u);
  }
  int mid = (l + r) / 2;
  solve(l, mid, plecak, nowe);
  solve(mid, r, plecak, nowe);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n >> maxMass;
  pot[0] = 1;
  const long long p = 1e7 + 19;
  for (int i = 1; i <= maxMass; ++i) pot[i] = (pot[i - 1] * p) % MOD;
  vector<int> dane(n);
  vector<long long> plecak(maxMass + 1);
  int id = 0;
  for (int i = 0; i < n; ++i) {
    in[id] = 0;
    out[id] = N;
    cin >> val[id] >> mass[id];
    dane[id] = id;
    ++id;
  }
  int q;
  cin >> q;
  int last = 0;
  while (q--) {
    int t;
    cin >> t;
    if (t == 3) {
      ++last;
    } else if (t == 2) {
      int x;
      cin >> x;
      out[x - 1] = last + 1;
    } else {
      int v, w;
      cin >> v >> w;
      dane.push_back(id);
      val[id] = v;
      mass[id] = w;
      in[id] = last;
      out[id] = N;
      ++id;
    }
  }
  solve(0, last, plecak, dane);
  return 0;
}
