Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m, nn, Q, t;
int w[40000], v[40000], wen[40000], f[40000], g[1010], a[40000 * 4], id[40000];
vector<int> tr[40000 * 4];
void build(int q, int x, int y, int l, int r, int k) {
  if (l <= x && y <= r) {
    tr[q].push_back(k);
    return;
  }
  int mid = (x + y) / 2;
  if (l <= mid) build(q * 2, x, mid, l, r, k);
  if (mid < r) build(q * 2 + 1, mid + 1, y, l, r, k);
}
void Build(int q, int x, int y) {
  if (x == y) {
    a[q] = wen[x];
    return;
  }
  int mid = (x + y) / 2;
  Build(q * 2, x, mid);
  Build(q * 2 + 1, mid + 1, y);
  a[q] = a[q * 2] | a[q * 2 + 1];
}
void jia(int a[1010], int y, int x) {
  int i;
  for (i = m; i >= x; i--) a[i] = max(a[i], a[i - x] + y);
}
void print() {
  int i, an = 0;
  for (i = m; i; i--) an = (1ll * 10000019 * an + g[i]) % 1000000007;
  printf("%d\n", an);
}
void sao(int q, int x, int y) {
  if (!a[q]) return;
  int mid = (x + y) / 2, vv[40000], i;
  for (i = 0; i <= m; i++) vv[i] = g[i];
  for (auto j : tr[q]) jia(g, v[j], w[j]);
  if (x == y) {
    print();
    for (i = 0; i <= m; i++) g[i] = vv[i];
    return;
  }
  sao(q * 2, x, mid);
  sao(q * 2 + 1, mid + 1, y);
  for (i = 0; i <= m; i++) g[i] = vv[i];
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d%d", &v[i], &w[i]), f[i] = 1, id[i] = i;
  scanf("%d", &Q);
  nn = n + Q;
  t = n;
  for (i = 1; i <= Q; i++) {
    int _;
    scanf("%d", &_);
    if (_ == 1) {
      ++t;
      scanf("%d%d", &v[t], &w[t]), f[t] = 1;
      id[t] = n + i;
    }
    if (_ == 2) {
      int x;
      scanf("%d", &x);
      build(1, 1, nn, id[x], i + n, x);
      f[x] = 0;
    }
    if (_ == 3) wen[i + n] = 1;
  }
  for (i = 1; i <= n; i++)
    if (f[i]) jia(g, v[i], w[i]);
  for (i = n + 1; i <= t; i++)
    if (f[i]) build(1, 1, nn, id[i], nn, i);
  Build(1, 1, nn);
  sao(1, 1, nn);
  return 0;
}