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;
int weight[36206], value[36206];
int st[36206], ep[36206];
vector<int> ask;
struct state {
  int dp[1002];
  void clear() { memset(dp, 0, sizeof dp); }
  void modi(int w, int h) {
    for (int i = 1002 - w - 1; i >= 0; i--) {
      dp[i + w] = max(dp[i + w], dp[i] + h);
    }
  }
  void modi(vector<pair<int, int> > v) {
    for (int i = 0; i < v.size(); i++) modi(v[i].first, v[i].second);
  }
} o;
state base;
vector<state> road;
vector<int> roadid;
int lc[36206 * 2], rc[36206 * 2];
vector<pair<int, int> > ops[36206 * 2];
int tot;
void modi(int a, int b, int l, int r, int &id, int &w, int &h) {
  if (!id) id = ++tot;
  if (l >= a and r <= b) {
    ops[id].push_back(make_pair(w, h));
    return;
  }
  if (a <= (l + r) / 2) modi(a, b, l, (l + r) / 2, lc[id], w, h);
  if (b > (l + r) / 2) modi(a, b, (l + r) / 2 + 1, r, rc[id], w, h);
}
vector<int> newid;
void query(int x, int l, int r, int id) {
  if (!id) return;
  newid.push_back(id);
  if (x <= (l + r) / 2)
    query(x, l, (l + r) / 2, lc[id]);
  else
    query(x, (l + r) / 2 + 1, r, rc[id]);
}
int main() {
  scanf("%d%d", &n, &k);
  int zip = 0;
  for (int i = 1; i <= n; i++) {
    st[++zip] = 1;
    scanf("%d%d", &value[zip], &weight[zip]);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int ops;
    scanf("%d", &ops);
    if (ops == 1) {
      st[++zip] = i;
      scanf("%d%d", &value[zip], &weight[zip]);
    } else if (ops == 2) {
      int x;
      scanf("%d", &x);
      ep[x] = i;
    } else {
      ask.push_back(i);
    }
  }
  int e = 0;
  for (int i = 1; i <= zip; i++)
    if (st[i]) {
      if (!ep[i]) ep[i] = 36206;
      modi(st[i], ep[i], 1, 36206, e, weight[i], value[i]);
    }
  for (int i = 0; i < ask.size(); i++) {
    newid.clear();
    int x = ask[i];
    query(ask[i], 1, 36206, e);
    int st = 0;
    for (int i = 0; i < newid.size() and i < roadid.size(); i++)
      if (newid[i] == roadid[i]) st = i + 1;
    while (roadid.size() > st) roadid.pop_back(), road.pop_back();
    for (int i = st; i < newid.size(); i++) {
      int id = newid[i];
      roadid.push_back(id);
      if (road.size())
        road.push_back(road[road.size() - 1]);
      else
        road.push_back(base);
      road[road.size() - 1].modi(ops[id]);
    }
    assert(road.size() == newid.size());
    long long ek = 1, ans = 0;
    for (int i = 1; i <= k; i++, ek = ek * 10000019 % 1000000007)
      ans += road[road.size() - 1].dp[i] * ek % 1000000007;
    printf("%lld\n", ans % 1000000007);
  }
  return 0;
}
