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 MAXM = 3e4 + 10;
const int MAXW = 1e3 + 10;
const long long M1 = 1e7 + 19;
const long long M2 = 1e9 + 7;
struct item {
  int s, t, w;
  long long v;
};
int W, m, cnt = 0;
bool bq[MAXM];
long long pw[MAXW], f[20][MAXW];
item r[MAXM];
vector<int> st[MAXM * 20];
void lucky() {
  cin.tie(0);
  cin >> cnt >> W;
  for (int i = 1; i <= cnt; i++) cin >> r[i].v >> r[i].w, r[i].s = 1;
  cin >> m;
}
long long calc(long long *f) {
  long long res = 0;
  for (int i = 1; i <= W; i++) res = (res + 1LL * f[i] * pw[i - 1]) % M2;
  return res;
}
void solve(int ql, int qr, int o, int d) {
  int mid = (ql + qr) >> 1;
  for (int i = 1; i <= W; i++) f[d][i] = f[d - 1][i];
  if (ql < qr)
    for (int i = 0; i < st[o].size(); i++) {
      int x = st[o][i];
      if (r[x].s <= ql && qr <= r[x].t)
        for (int j = W; j >= r[x].w; j--)
          f[d][j] = max(f[d][j], f[d][j - r[x].w] + r[x].v);
      else {
        if (r[x].s <= mid) st[((o) << 1)].push_back(x);
        if (r[x].t > mid) st[((o) << 1 | 1)].push_back(x);
      }
    }
  else {
    for (int i = 0; i < st[o].size(); i++) {
      int x = st[o][i];
      if (r[x].s <= ql && qr <= r[x].t)
        for (int j = W; j >= r[x].w; j--)
          f[d][j] = max(f[d][j], f[d][j - r[x].w] + r[x].v);
    }
    if (bq[ql]) cout << calc(f[d]) << endl;
    return;
  }
  solve(ql, mid, ((o) << 1), d + 1);
  solve(mid + 1, qr, ((o) << 1 | 1), d + 1);
}
int main() {
  int o, x;
  lucky();
  for (int i = 1; i <= cnt; i++) r[i].t = m;
  for (int i = 1; i <= m; i++)
    if (cin >> o, o == 1)
      cnt++, r[cnt].s = i, r[cnt].t = m, cin >> r[cnt].v >> r[cnt].w;
    else if (o == 2)
      cin >> x, r[x].t = i - 1;
    else
      bq[i] = 1;
  pw[0] = 1;
  for (int i = 1; i <= W; i++) pw[i] = (pw[i - 1] * M1) % M2;
  for (int i = 1; i <= cnt; i++) st[1].push_back(i);
  solve(1, m, 1, 1);
  return 0;
}
