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 N = 1e5;
const long long INF = 1e9 + 19;
const int MOD = 1e9 + 7;
const long long p = 1e7 + 19;
const int K = 1111;
struct Query {
  int type, c, w, id;
  Query() {}
  Query(int type, int c, int w, int id) : type(type), c(c), w(w), id(id) {}
};
struct Item {
  int w, c;
  Item() {}
  Item(int w, int c) : w(w), c(c) {}
};
int n, k;
int when[N];
vector<Query> q;
vector<Item> data[N * 2];
int answer[N];
int cur;
int dp[K];
void read() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int c, w;
    scanf("%d%d", &c, &w);
    q.push_back(Query(1, c, w, i));
  }
  int cnt = 0;
  scanf("%d", &cnt);
  for (int i = 0; i < cnt; i++) {
    int type;
    scanf("%d", &type);
    if (type == 3)
      q.push_back(Query(3, -1, -1, cur++));
    else if (type == 1) {
      int c, w;
      scanf("%d%d", &c, &w);
      q.push_back(Query(1, c, w, n));
      n++;
    } else if (type == 2) {
      int id;
      scanf("%d", &id);
      id--;
      q.push_back(Query(2, -1, -1, id));
    }
  }
}
void add(int v, int ll, int rr, int l, int r, Item item) {
  if (l >= rr || ll >= r) return;
  if (l <= ll && rr <= r) {
    data[v].push_back(item);
    return;
  }
  add(v * 2 + 1, ll, (ll + rr) / 2, l, r, item);
  add(v * 2 + 2, (ll + rr) / 2, rr, l, r, item);
}
void go(int v, int ll, int rr) {
  vector<int> tmp(k + 1);
  for (int i = 0; i <= k; i++) tmp[i] = dp[i];
  for (auto x : data[v]) {
    for (int j = k - x.w; j >= 0; j--)
      dp[j + x.w] = max(dp[j + x.w], dp[j] + x.c);
  }
  if (ll + 1 == rr) {
    if (q[ll].type == 3) {
      int pp = 1;
      int sum = 0;
      for (int i = 1; i <= k; i++) {
        sum = (sum + dp[i] * 1ll * pp) % MOD;
        pp = (pp * 1ll * p) % MOD;
      }
      answer[q[ll].id] = sum;
    }
    for (int i = 0; i <= k; i++) dp[i] = tmp[i];
    return;
  }
  go(v * 2 + 1, ll, (ll + rr) / 2);
  go(v * 2 + 2, (ll + rr) / 2, rr);
  for (int i = 0; i <= k; i++) dp[i] = tmp[i];
}
void solve() {
  set<int> g;
  for (auto x : q) {
    if (x.type == 1) g.insert(x.id);
    if (x.type == 2) {
      assert(g.count(x.id));
      g.erase(x.id);
    }
  }
  for (auto x : g) q.push_back(Query(2, -1, -1, x));
  for (int i = 0; i < (int)q.size(); i++) {
    if (q[i].type == 1) when[q[i].id] = i;
    if (q[i].type == 2) {
      int l = when[q[i].id];
      int r = i;
      add(0, 0, q.size(), l, r, Item(q[l].w, q[l].c));
    }
  }
  go(0, 0, q.size());
  for (int i = 0; i < cur; i++) printf("%d\n", answer[i]);
}
void printAns() {}
void stress() {}
int main() {
  if (1) {
    int k = 1;
    for (int tt = 0; tt < k; tt++) {
      read();
      solve();
      printAns();
    }
  } else {
    stress();
  }
  return 0;
}
