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 = 1005;
const int MAXM = 30005;
const long long P = 10000019;
const long long Q = 1000000007;
int n, k, q, tot, lft[MAXM], rit[MAXM];
long long f[MAXN], pw[MAXN];
bool is_q[MAXM];
struct Object {
  int v, w;
} obj[MAXM];
vector<Object> vec[MAXM * 4];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    (x *= 10) += ch - '0';
    ch = getchar();
  }
  return x * f;
}
void upd(int p, int l, int r, int ql, int qr, Object o) {
  if (ql <= l && r <= qr) return vec[p].push_back(o), void();
  int mid = (l + r) >> 1;
  if (ql <= mid) upd(p << 1, l, mid, ql, qr, o);
  if (qr > mid) upd(p << 1 | 1, mid + 1, r, ql, qr, o);
}
void solve(int p, int l, int r) {
  for (Object o : vec[p])
    for (int i = k; i >= o.w; i--) f[i] = max(f[i], f[i - o.w] + o.v);
  if (l == r) {
    if (is_q[l]) {
      long long ans = 0;
      for (int i = 1; i <= k; i++) (ans += f[i] * pw[i - 1]) %= Q;
      printf("%lld\n", ans);
    }
    return;
  }
  long long cpy[MAXN] = {0};
  int mid = (l + r) >> 1;
  memcpy(cpy, f, sizeof(cpy));
  solve(p << 1, l, mid);
  memcpy(f, cpy, sizeof(f));
  solve(p << 1 | 1, mid + 1, r);
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++)
    tot++, obj[tot].v = read(), obj[tot].w = read(), lft[tot] = 1;
  q = read();
  for (int i = 1, opt, x, y; i <= q; i++) {
    opt = read();
    if (opt == 1)
      tot++, obj[tot].v = read(), obj[tot].w = read(), lft[tot] = i,
             rit[tot] = q;
    if (opt == 2) rit[read()] = i;
    if (opt == 3) is_q[i] = true;
  }
  for (int i = 1; i <= tot; i++)
    if (!rit[i]) rit[i] = q;
  for (int i = 1; i <= tot; i++) upd(1, 1, q, lft[i], rit[i], obj[i]);
  pw[0] = 1;
  for (int i = 1; i <= k; i++) pw[i] = pw[i - 1] * P % Q;
  solve(1, 1, q);
  return 0;
}
