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;
template <class T>
inline void read(T& x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
const int N = 30000 + 5;
const int M = 1000 + 5;
const int P = 10000000 + 19;
const int mo = 1000000000 + 7;
int n, k, q, top;
int vis[N];
long long p[M];
int v[N << 1], w[N << 1], st[N << 1], ed[N << 1];
int ans[20][M];
std::vector<int> g[N << 2];
void modify(int x, int l, int r, int s, int t, int id) {
  if (s <= l && r <= t) {
    g[x].push_back(id);
    return;
  }
  int m = (l + r) >> 1;
  if (s <= m) modify((x << 1), l, m, s, t, id);
  if (m < t) modify((x << 1 | 1), m + 1, r, s, t, id);
}
void solve(int x, int l, int r) {
  int* f = &ans[++top][0];
  copy(ans[top - 1], ans[top - 1] + M, f);
  for (int i = 0; i < g[x].size(); i++) {
    int id = g[x][i];
    for (int j = k; j >= w[id]; j--) f[j] = max(f[j], f[j - w[id]] + v[id]);
  }
  if (l == r) {
    if (vis[l]) {
      long long tmp = 0;
      for (int i = 1; i <= k; i++) (tmp += f[i] * p[i - 1] % mo) %= mo;
      printf("%lld\n", tmp);
    }
    top--;
    return;
  }
  int m = (l + r) >> 1;
  solve((x << 1), l, m);
  solve((x << 1 | 1), m + 1, r);
  top--;
}
int main() {
  read(n), read(k);
  for (int i = 1; i <= n; i++) read(v[i]), read(w[i]), st[i] = 1;
  read(q);
  for (int i = 1, op, x; i <= q; i++) {
    read(op);
    if (op == 3) vis[i] = 1;
    if (op == 1) {
      n++;
      read(v[n]), read(w[n]);
      st[n] = i;
    }
    if (op == 2) {
      read(x);
      ed[x] = i;
    }
  }
  p[0] = 1;
  for (int i = 1; i <= k; i++) p[i] = p[i - 1] * P % mo;
  for (int i = 1; i <= n; i++) modify(1, 1, q, st[i], ed[i] ? ed[i] : q, i);
  solve(1, 1, q);
  return 0;
}
