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 long long mod = 1000000007;
int n, m, len, q, b[30005];
long long hea[10000000], wen[10000000], nex[10000000], f[20][1005], c[10000];
struct arr {
  long long bo, v, w, l, r;
} a[40005];
void add(int l, int r, int k, int x) {
  if (a[x].l <= l && a[x].r >= r) {
    ++len;
    nex[len] = hea[k];
    wen[len] = x;
    hea[k] = len;
    return;
  }
  int mid = (l + r) >> 1;
  if (a[x].l <= mid) add(l, mid, k * 2, x);
  if (a[x].r > mid) add(mid + 1, r, k * 2 + 1, x);
}
void find(int l, int r, int k, int t) {
  if (l == r) {
    if (!b[l]) return;
    for (int i = 0; i <= m; i++) f[t][i] = f[t - 1][i];
    for (int i = hea[k]; i; i = nex[i])
      for (int j = m; j >= a[wen[i]].w; j--)
        f[t][j] = max(f[t][j], f[t][j - a[wen[i]].w] + a[wen[i]].v);
    long long ans = 0;
    for (int i = 1; i <= m; i++) ans = (ans + f[t][i] % mod * c[i]) % mod;
    printf("%lld\n", ans);
    return;
  }
  for (int i = 0; i <= m; i++) f[t][i] = f[t - 1][i];
  for (int i = hea[k]; i; i = nex[i])
    for (int j = m; j >= a[wen[i]].w; j--)
      f[t][j] = max(f[t][j], f[t][j - a[wen[i]].w] + a[wen[i]].v);
  int mid = (l + r) >> 1;
  find(l, mid, k * 2, t + 1);
  find(mid + 1, r, k * 2 + 1, t + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &a[i].v, &a[i].w);
    a[i].l = 1;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int bo, x;
    scanf("%d", &bo);
    if (bo == 1) {
      ++n;
      scanf("%lld%lld", &a[n].v, &a[n].w);
      a[n].l = i;
    } else if (1 - (bo & 1)) {
      scanf("%d", &x);
      a[x].r = i - 1;
      if (a[x].r >= a[x].l) add(1, q, 1, x);
      a[x].bo = 1;
    } else
      b[i] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (!a[i].bo) a[i].r = q, add(1, q, 1, i);
  c[1] = 1;
  for (int i = 2; i <= m; i++) c[i] = c[i - 1] * 10000019 % mod;
  find(1, q, 1, 1);
  return 0;
}
