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 p = 1e7 + 19, mo = 1e9 + 7;
int n, k, T, cnt, tot, opt, x, v[30010 + 5000], w[30010 + 5000],
    lef[30010 + 5000], rig[30010 + 5000];
vector<int> wh[30010 * 4];
int f[1010];
void add(int o, int l, int r, int x, int L, int R) {
  if (r < L || R < l) return;
  if (L <= l && r <= R) return wh[o].push_back(x);
  int mid((l + r) >> 1);
  add(o << 1, l, mid, x, L, R);
  add(o << 1 | 1, mid + 1, r, x, L, R);
}
void copy(int *g, int *f) {
  for (int i = 1; i <= k; i++) g[i] = f[i];
}
void upd(int *f, int o) {
  for (int i = 0, s = wh[o].size(), x; i < s; i++) {
    x = wh[o][i];
    for (int j = k; j >= w[x]; j--)
      f[j] = ((f[j - w[x]] + v[x]) > (f[j]) ? f[j - w[x]] + v[x] : f[j]);
  }
}
void print(int *f) {
  int ans(0), s(1), t(0);
  for (int i = 1; i <= k; i++) {
    ans += 1LL * s * (t = ((f[i]) > (t) ? f[i] : t)) % mo, s = 1LL * s * p % mo;
    if (ans >= mo) ans -= mo;
  }
  printf("%d\n", ans);
}
void work(int o, int l, int r, int *f) {
  int nt[1010];
  copy(nt, f), upd(nt, o);
  if (l < r) {
    int mid((l + r) >> 1);
    work(o << 1, l, mid, nt);
    work(o << 1 | 1, mid + 1, r, nt);
  } else
    print(nt);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d %d", v + i, w + i), lef[i] = cnt + 1;
  scanf("%d", &T);
  memset(rig, -1, sizeof(rig));
  for (int i = 1; i <= T; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      tot++;
      scanf("%d %d", v + n + tot, w + n + tot);
      lef[n + tot] = cnt + 1;
    }
    if (opt == 2) {
      scanf("%d", &x);
      rig[x] = cnt;
    }
    if (opt == 3) cnt++;
  }
  for (int i = 1; i <= n + tot; i++)
    if (lef[i]) add(1, 1, cnt, i, lef[i], rig[i] > -1 ? rig[i] : cnt);
  work(1, 1, cnt, f);
  return 0;
}
