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 N = 3 * 1e4 + 10, K = 1010, p = 1e7 + 19, q = 1e9 + 7, LOG = 30;
int dp[LOG][K], n, k, st[N], en[N];
pair<int, int> d[2 * N];
long long power[K], ans[N];
;
vector<pair<int, int> > bz[4 * N];
void pre_p() {
  power[0] = 1;
  for (int i = 1; i <= k; i++) power[i] = (power[i - 1] * p) % q;
  return;
}
void add(int id, int L, int R, int s, int e, pair<int, int> value) {
  if (L <= s && e <= R) {
    bz[id].push_back(value);
    return;
  }
  if (L >= e || s >= R) return;
  int mid = (s + e) / 2;
  add(2 * id, L, R, s, mid, value);
  add(2 * id + 1, L, R, mid, e, value);
  return;
}
void build(int lg, int id, int s, int e) {
  if (lg > 0)
    for (int i = 0; i <= k; i++) dp[lg][i] = dp[lg - 1][i];
  for (auto X : bz[id]) {
    int val = X.first, mas = X.second;
    for (int i = k; i >= 1; i--) {
      if (i - mas < 0) continue;
      dp[lg][i] = max(dp[lg][i], dp[lg][i - mas] + val);
    }
  }
  if (e - s < 2) {
    for (int i = 1; i <= k; i++)
      ans[s] = (ans[s] + (power[i - 1] * dp[lg][i]) % q) % q;
    return;
  }
  int mid = (s + e) / 2;
  build(lg + 1, 2 * id, s, mid);
  build(lg + 1, 2 * id + 1, mid, e);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tim = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> d[i].first >> d[i].second;
    st[i] = 0;
    en[i] = N;
  }
  pre_p();
  int q, number = n;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int value, mas;
      cin >> value >> mas;
      d[number] = {value, mas};
      st[number] = tim;
      en[number] = N;
      number++;
    } else if (t == 2) {
      int id;
      cin >> id;
      id--;
      en[id] = tim;
    } else {
      tim++;
    }
  }
  for (int i = 0; i < number; i++) {
    en[i] = min(en[i], tim);
    add(1, st[i], en[i], 0, tim, d[i]);
  }
  build(0, 1, 0, tim);
  for (int i = 0; i < tim; i++) cout << ans[i] << '\n';
}
