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;
int n, k, q, ti[30005], vi, wi, xi, v[15005], w[15005], st[15005], en[15005];
vector<int> tree[4 * 30005];
int dp[30][1005];
void add(int i, int L, int R, int p, int q, int y) {
  if ((q < L) || (R < p)) return;
  if ((p <= L) && (R <= q)) {
    tree[i].push_back(y);
    return;
  }
  int mid = (L + R) / 2;
  add(2 * i, L, mid, p, q, y);
  add(2 * i + 1, mid + 1, R, p, q, y);
}
void solve(int i, int L, int R, int d) {
  for (int j = 0; j < tree[i].size(); j++)
    for (int z = k; z >= 0; z--)
      if (z >= w[tree[i][j]])
        dp[d][z] = max(dp[d][z], dp[d][z - w[tree[i][j]]] + v[tree[i][j]]);
  if (L == R) {
    if (ti[L] == 3) {
      long long r = 0, p = 1;
      for (int j = 1; j <= k; j++) {
        r += dp[d][j] * p;
        r %= 1000000007;
        p *= 10000019;
        p %= 1000000007;
      }
      printf("%I64d\n", r);
    }
    return;
  }
  int mid = (L + R) / 2;
  for (int z = 0; z <= k; z++) dp[d + 1][z] = dp[d][z];
  solve(2 * i, L, mid, d + 1);
  for (int z = 0; z <= k; z++) dp[d + 1][z] = dp[d][z];
  solve(2 * i + 1, mid + 1, R, d + 1);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d %d", &v[i], &w[i]);
  scanf("%d", &q);
  for (int i = 0; i < n; i++) {
    st[i] = 0;
    en[i] = q - 1;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d", &ti[i]);
    if (ti[i] == 1) {
      scanf("%d %d", &vi, &wi);
      v[n] = vi, w[n] = wi;
      st[n] = i;
      en[n] = q - 1;
      n++;
    } else if (ti[i] == 2) {
      scanf("%d", &xi);
      en[xi - 1] = i;
    } else
      continue;
  }
  for (int i = 0; i < n; i++) add(1, 0, q - 1, st[i], en[i], i);
  solve(1, 0, q - 1, 0);
  return 0;
}
