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 n, q, k, cnt, s[30010], t[30010], w[30010];
long long v[30010], pw[1010], ans[30010 * 3], dp[1010];
bool qry[30010];
vector<int> tr[30010 << 2];
void modify(int p, int L, int R, int l, int r, int id) {
  if (l <= L && R <= r) {
    tr[p].push_back(id);
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) modify(p << 1, L, mid, l, r, id);
  if (r > mid) modify(p << 1 | 1, mid + 1, R, l, r, id);
}
void solve(int p, int L, int R) {
  long long tmp[1010];
  for (int i = 1; i <= k; i++) tmp[i] = dp[i];
  for (int i : tr[p])
    for (int j = k; j >= w[i]; j--) dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
  if (L == R) {
    if (qry[L]) {
      for (int i = 1; i <= k; i++) {
        ans[L] = ans[L] + dp[i] * pw[i - 1] % 1000000007;
        if (ans[L] >= 1000000007) ans[L] -= 1000000007;
      }
    }
  } else {
    int mid = (L + R) >> 1;
    solve(p << 1, L, mid);
    solve(p << 1 | 1, mid + 1, R);
  }
  for (int i = 1; i <= k; i++) dp[i] = tmp[i];
}
int main() {
  scanf("%d%d", &n, &k);
  pw[0] = 1;
  for (int i = 1; i <= k; i++) pw[i] = pw[i - 1] * 10000019 % 1000000007;
  for (int i = 1; i <= n; i++) {
    scanf("%lld%d", &v[i], &w[i]);
    s[i] = 1;
  }
  cnt = n;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int op, id;
    scanf("%d", &op);
    if (op == 1) {
      cnt++;
      scanf("%lld%d", &v[cnt], &w[cnt]);
      s[cnt] = i;
    }
    if (op == 2) {
      scanf("%d", &id);
      t[id] = i;
    }
    if (op == 3) qry[i] = 1;
  }
  for (int i = 1; i <= cnt; i++) {
    if (!t[i]) t[i] = q;
    modify(1, 1, q, s[i], t[i], i);
  }
  solve(1, 1, q);
  for (int i = 1; i <= q; i++)
    if (qry[i]) printf("%lld\n", ans[i]);
  return 0;
}