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;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 3e4 + 10;
const int K = 1e3 + 2;
const int P = 1e7 + 19;
const int Q = 1e9 + 7;
const long double eps = 1e-12;
int n, k, q, v[N], w[N], st[N], ft[N], m;
long long ans[N];
vector<int> s[4 * N];
vector<int> ask;
vector<vector<int> > knap;
void SET(int ql, int qr, int ind, int l = 0, int r = m, int idx = 1) {
  if (r <= ql || qr <= l) return;
  if (ql <= l && r <= qr) {
    s[idx].push_back(ind);
    return;
  }
  int mid = (l + r) / 2;
  SET(ql, qr, ind, l, mid, idx * 2 + 0);
  SET(ql, qr, ind, mid, r, idx * 2 + 1);
}
void COUNT(int j) {
  vector<int> &vec = knap.back();
  long long res = 0, tav = 1, d = 0;
  for (int i = 1; i <= k; i++) {
    d = max(d, (long long)vec[i]);
    res += (1LL * d * tav) % Q;
    tav = (1LL * tav * P) % Q;
  }
  ans[j] = res % Q;
}
void add(int v, int w) {
  vector<int> &pp = knap.back();
  vector<int> vec;
  for (int i = 0; i < pp.size(); i++) vec.push_back(pp[i]);
  for (int i = K - 1; i >= w; i--) vec[i] = max(vec[i - w] + v, vec[i]);
  knap.push_back(vec);
}
void dfs(int l = 0, int r = m, int idx = 1) {
  for (int i = 0; i < s[idx].size(); i++) add(v[s[idx][i]], w[s[idx][i]]);
  if (r - l < 2) {
    COUNT(l);
    for (int i = 0; i < s[idx].size(); i++) knap.pop_back();
    return;
  }
  int mid = (l + r) / 2;
  dfs(l, mid, idx * 2 + 0);
  dfs(mid, r, idx * 2 + 1);
  for (int i = 0; i < s[idx].size(); i++) knap.pop_back();
}
int main() {
  vector<int> vec;
  vec.resize(K);
  knap.push_back(vec);
  n = in(), k = in();
  for (int i = 1; i <= n; i++) v[i] = in(), w[i] = in(), st[i] = 0;
  q = in();
  for (int i = 0; i < N; i++) ft[i] = q;
  n++;
  for (int i = 0; i < q; i++) {
    int t = in();
    if (t == 1)
      v[n] = in(), w[n] = in(), st[n] = i, n++;
    else if (t == 2) {
      int x = in();
      ft[x] = i + 1;
    } else if (t == 3)
      ask.push_back(i);
  }
  m = q;
  for (int i = 1; i < n; i++) SET(st[i], ft[i], i);
  dfs();
  for (int i = 0; i < ask.size(); i++) cout << ans[ask[i]] << "\n";
  return 0;
}
