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;
struct unit {
  int w, v, l, r;
};
struct unit a[15010];
struct unit2 {
  int opt, x, y;
};
struct unit2 q[30010];
int n, m, nq, tot, ans, np = 10000019, mod = 1000000007;
int fp[1010];
long long f[20][1010];
vector<int> segt[120010];
vector<int> rec;
int get_node() {
  int ret;
  if (!rec.empty()) {
    ret = rec.back();
    rec.pop_back();
  } else {
    tot++;
    ret = tot;
  }
  return ret;
}
void calc(int k) {
  int i;
  for (i = 1; i <= m; i++)
    ans = (ans + (long long)f[k][i] % mod * fp[i - 1] % mod) % mod;
}
void change(int i, int l, int r, int x, int y, int z) {
  if (x <= l && y >= r) {
    segt[i].push_back(z);
    return;
  }
  if (x > r || y < l) return;
  change(i * 2, l, (l + r) / 2, x, y, z);
  change(i * 2 + 1, (l + r) / 2 + 1, r, x, y, z);
}
void query(int i, int l, int r, int k) {
  int j, p, i1;
  if (l > r) return;
  p = get_node();
  for (j = 0; j <= m; j++) f[p][j] = f[k][j];
  for (j = 0; j < segt[i].size(); j++)
    for (i1 = m; i1 >= a[segt[i][j]].w; i1--)
      f[p][i1] = max(f[p][i1], f[p][i1 - a[segt[i][j]].w] + a[segt[i][j]].v);
  if (l == r) {
    if (q[l].opt == 3) {
      ans = 0;
      calc(p);
      printf("%d\n", ans);
    }
    rec.push_back(p);
    return;
  }
  query(i * 2, l, (l + r) / 2, p);
  query(i * 2 + 1, (l + r) / 2 + 1, r, p);
  rec.push_back(p);
}
int main() {
  int i, j;
  scanf("%d%d", &n, &m);
  fp[0] = 1;
  for (i = 1; i <= m; i++) fp[i] = (long long)fp[i - 1] * np % mod;
  for (i = 1; i <= n; i++) scanf("%d%d", &a[i].v, &a[i].w);
  scanf("%d", &nq);
  for (i = 1; i <= nq; i++) {
    scanf("%d", &q[i].opt);
    if (q[i].opt == 1) {
      scanf("%d%d", &q[i].x, &q[i].y);
      n++;
      a[n].v = q[i].x;
      a[n].w = q[i].y;
      a[n].l = i;
    } else if (q[i].opt == 2) {
      scanf("%d", &q[i].x);
      a[q[i].x].r = i;
    }
  }
  for (i = 1; i <= n; i++) {
    if (a[i].l == 0) a[i].l = 0;
    if (a[i].r == 0) a[i].r = nq + 1;
  }
  for (i = 1; i <= n; i++) change(1, 1, nq, a[i].l + 1, a[i].r - 1, i);
  query(1, 1, nq, 0);
  return 0;
}
