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, Q;
const int N_ = 5050, K_ = 1050, Q_ = 30050, Q1_ = 30050;
const int lgQ_ = 20;
struct exhibit {
  int v, w, l, r;
  exhibit(int v = 0, int w = 0, int l = 0, int r = 0)
      : v(v), w(w), l(l), r(r) {}
};
exhibit exhibits[N_ + Q1_];
vector<int> time3;
long long ans[Q_];
bool check_time3[Q_];
const long long hashP = (long long)1e7 + 19, hashQ = (long long)1e9 + 7;
int tb[lgQ_][K_];
void solve(int l, int r, vector<exhibit> &cur, int level) {
  for (int m = 0; m <= K; m++) tb[level][m] = tb[level - 1][m];
  vector<exhibit> nxt;
  for (exhibit e : cur) {
    if (e.l <= l && r <= e.r) {
      for (int i = K; i >= e.w; i--)
        tb[level][i] = max(tb[level][i], tb[level][i - e.w] + e.v);
    } else if (e.l <= r && l <= e.r) {
      nxt.push_back(e);
    }
  }
  if (l == r) {
    if (check_time3[l]) {
      long long mul = 1;
      for (int m = 1; m <= K; m++) {
        ans[l] = (ans[l] + tb[level][m] * mul) % hashQ;
        mul = (mul * hashP) % hashQ;
      }
    }
    return;
  } else {
    int mid = (l + r) >> 1;
    solve(l, mid, nxt, level + 1);
    solve(mid + 1, r, nxt, level + 1);
  }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    exhibit &e = exhibits[i];
    scanf("%d%d", &e.v, &e.w);
    e.l = 1;
  }
  scanf("%d", &Q);
  ++Q;
  for (int q = 2; q <= Q; q++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      exhibit &e = exhibits[++N];
      scanf("%d%d", &e.v, &e.w);
      e.l = q;
    } else if (t == 2) {
      int i;
      scanf("%d", &i);
      exhibits[i].r = q;
    } else if (t == 3) {
      time3.push_back(q);
      check_time3[q] = true;
    }
  }
  ++Q;
  for (int i = 1; i <= N; i++) {
    exhibit &e = exhibits[i];
    if (e.r == 0) e.r = Q;
  }
  vector<exhibit> cur;
  for (int i = 1; i <= N; i++) cur.push_back(exhibits[i]);
  solve(1, Q, cur, 1);
  for (auto t : time3) printf("%lld\n", ans[t]);
  return 0;
}
