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 = 40000, maxm = 2100000, maxk = 1013;
int n, k, Q, w[maxn], v[maxn], ins[maxn], del[maxn], ty, x;
bool isq[maxn];
vector<int> tag[maxm];
void modify(int t, int l, int r, int x, int y, int o) {
  if (l >= x && r <= y) {
    tag[t].push_back(o);
    return;
  }
  if (x <= ((l + r) >> 1)) modify((t << 1), l, ((l + r) >> 1), x, y, o);
  if (y > ((l + r) >> 1))
    modify(((t << 1) | 1), ((l + r) >> 1) + 1, r, x, y, o);
}
void ask(int t, int l, int r, int s[]) {
  int st[1010];
  for (int i = 1; i <= k; i++) st[i] = s[i];
  for (int i = 0; i < (int)tag[t].size(); i++) {
    int o = tag[t][i];
    for (int j = k; j >= w[o]; j--) st[j] = max(st[j], st[j - w[o]] + v[o]);
  }
  if (l == r) {
    if (isq[l]) {
      int ans = 0;
      for (int i = 1, pp = 1; i <= k;
           i++, pp = (long long)pp * 10000019 % 1000000007)
        ans = (ans + (long long)pp * st[i] % 1000000007) % 1000000007;
      printf("%d\n", ans);
    }
    return;
  }
  ask((t << 1), l, ((l + r) >> 1), st);
  ask(((t << 1) | 1), ((l + r) >> 1) + 1, r, st);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", v + i, w + i);
    ins[i] = i;
  }
  scanf("%d", &Q);
  int tim = n + 1;
  for (int i = 1; i <= Q; i++, tim++) {
    scanf("%d", &ty);
    if (ty == 1) {
      n++;
      scanf("%d%d", v + n, w + n);
      ins[n] = tim;
    } else if (ty == 2) {
      scanf("%d", &x);
      del[x] = tim;
    } else
      isq[tim] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (!del[i]) del[i] = tim;
  for (int i = 1; i <= n; i++) modify(1, 1, tim, ins[i], del[i], i);
  int s[1010];
  memset(s, 0, sizeof(s));
  ask(1, 1, tim, s);
  return 0;
}
