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 maxn = 41000;
const int p = 1e7 + 19;
const int q = 1e9 + 7;
int ed[maxn], st[maxn];
pair<int, int> item[maxn];
bool query[maxn];
int pro[maxn];
int k;
vector<int> add[maxn * 4];
void update(int l, int r, int rt, int tl, int tr, int node) {
  if (tl <= l && r <= tr) {
    add[rt].push_back(node);
    return;
  }
  int mid = (l + r) >> 1;
  if (tl <= mid) update(l, mid, rt << 1, tl, tr, node);
  if (tr >= mid + 1) update(mid + 1, r, rt << 1 | 1, tl, tr, node);
}
long long ksm(long long num, long long t) {
  long long res = 1;
  num %= q;
  for (; t; t >>= 1, num = num * num % q) {
    if (t & 1) res = res * num % q;
  }
  return res;
}
int tmp[1005];
void solve(int l, int r, int rt) {
  int buc[1005] = {0};
  for (int i = (0), iend = ((int)add[rt].size() - 1); i <= iend; i++) {
    int id = add[rt][i];
    for (int j = k; j >= 0; j--) {
      if (j + item[id].second <= k)
        tmp[j + item[id].second] =
            max(tmp[j + item[id].second], tmp[j] + item[id].first);
    }
  }
  if (l == r) {
    if (query[l]) {
      long long ans = 0;
      for (int m = (1), mend = (k); m <= mend; m++) {
        ans += 1ll * tmp[m] * pro[m - 1] % q;
        ans %= q;
      }
      printf("%lld\n", ans);
    }
    return;
  }
  int mid = (l + r) >> 1;
  memcpy(buc, tmp, sizeof(tmp));
  solve(l, mid, rt << 1);
  memcpy(tmp, buc, sizeof(tmp));
  solve(mid + 1, r, rt << 1 | 1);
}
int main() {
  for (int i = (0), iend = (1005); i <= iend; i++) pro[i] = ksm(p, i);
  int n;
  scanf("%d%d", &n, &k);
  for (int i = (1), iend = (n); i <= iend; i++) {
    scanf("%d%d", &item[i].first, &item[i].second);
    st[i] = 1;
  }
  int q;
  scanf("%d", &q);
  int item_cnt = n;
  for (int i = (1), iend = (q); i <= iend; i++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      ++item_cnt;
      scanf("%d%d", &item[item_cnt].first, &item[item_cnt].second);
      st[item_cnt] = i;
    } else if (ty == 2) {
      int x;
      scanf("%d", &x);
      ed[x] = i;
    } else
      query[i] = 1;
  }
  for (int i = (1), iend = (item_cnt); i <= iend; i++)
    if (!ed[i]) ed[i] = q;
  for (int i = (1), iend = (item_cnt); i <= iend; i++)
    update(1, q, 1, st[i], ed[i], i);
  solve(1, q, 1);
  return 0;
}
