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;
long long base = 10000019;
long long mod = 1000000007;
int cst[15000];
int val[15000];
int ord[15000];
int fin[15000];
int dp[32768 * 2][1001];
int cdp[1001];
class segtree {
 public:
  vector<int> seg[32768 * 2];
  void add(int beg, int end, int node, int lb, int ub, int num) {
    if (ub < beg || end < lb) return;
    if (beg <= lb && ub <= end) {
      seg[node].push_back(num);
      return;
    }
    add(beg, end, node * 2, lb, (lb + ub) / 2, num);
    add(beg, end, node * 2 + 1, (lb + ub) / 2 + 1, ub, num);
  }
  void dfs(int node) {
    for (int i = 0; i < seg[node].size(); i++) {
      for (int j = 1; j <= 1000; j++) {
        cdp[j] = max(dp[node][j], cdp[j - 1]);
        if (j >= cst[seg[node][i]])
          cdp[j] =
              max(cdp[j], dp[node][j - cst[seg[node][i]]] + val[seg[node][i]]);
      }
      for (int j = 0; j <= 1000; j++) {
        dp[node][j] = cdp[j];
        cdp[j] = 0;
      }
    }
    if (node >= 32768) return;
    for (int j = 0; j <= 1000; j++) {
      dp[node * 2][j] = dp[node * 2 + 1][j] = dp[node][j];
    }
    dfs(node * 2);
    dfs(node * 2 + 1);
  }
};
segtree tree;
int main() {
  int num, gen;
  scanf("%d%d", &num, &gen);
  for (int i = 0; i < num; i++) {
    scanf("%d%d", &val[i], &cst[i]);
    ord[i] = 0;
  }
  fill(fin, fin + 15000, 32768 - 1);
  int pt = num;
  int query;
  scanf("%d", &query);
  vector<int> que;
  for (int i = 0; i < query; i++) {
    int z;
    scanf("%d", &z);
    if (z == 1) {
      int za, zb;
      scanf("%d%d", &za, &zb);
      val[pt] = za;
      cst[pt] = zb;
      ord[pt] = i + 1;
      pt++;
    }
    if (z == 2) {
      int za;
      scanf("%d", &za);
      za--;
      fin[za] = i + 1;
    }
    if (z == 3) que.push_back(i + 1);
  }
  for (int i = 0; i < pt; i++) tree.add(ord[i], fin[i], 1, 0, 32768 - 1, i);
  tree.dfs(1);
  for (int i = 0; i < que.size(); i++) {
    long long now = 0;
    for (int j = gen; j >= 1; j--) {
      now *= base;
      now += dp[que[i] + 32768][j];
      now %= mod;
    }
    printf("%I64d\n", now);
  }
}
