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 linf = 1e18 + 5;
const int mod = (int)1e9 + 7;
const int p = (int)1e7 + 19;
const int logN = 17;
const int inf = 1e9;
const int N = 40000;
int start[N], k, finish[N], t, x, y, z, n, m, dp[N], temp[N], q, Q[N];
vector<pair<int, int> > DT[N << 2];
pair<int, int> a[N];
void add(int k, int bas, int son, int x, int y, int ind) {
  if (bas > y || son < x) return;
  if (x <= bas && son <= y) {
    DT[k].push_back(a[ind]);
    return;
  }
  add((k + k), bas, (bas + son >> 1), x, y, ind);
  add((k + k + 1), (bas + son >> 1) + 1, son, x, y, ind);
}
void solve(int k, int bas, int son) {
  if (bas > son) return;
  vector<int> temp;
  temp.resize(::k + 1, 0);
  for (int i = 0; i <= ::k; i++) temp[i] = dp[i];
  for (__typeof(DT[k].begin()) it = DT[k].begin(); it != DT[k].end(); it++) {
    for (int i = ::k; i >= it->first; i--) {
      dp[i] = max(dp[i], dp[i - it->first] + it->second);
    }
  }
  if (bas == son) {
    if (Q[bas]) {
      long long ans = 0;
      int t = 1;
      for (int i = 1; i <= ::k; i++) {
        ans = (ans + t * (long long)dp[i] % mod) % mod;
        t = (long long)t * p % mod;
      }
      printf("%lld\n", ans);
    }
    for (int i = 0; i <= ::k; i++) dp[i] = temp[i];
    return;
  }
  solve((k + k), bas, (bas + son >> 1));
  solve((k + k + 1), (bas + son >> 1) + 1, son);
  for (int i = 0; i <= ::k; i++) dp[i] = temp[i];
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &x, &y);
    a[i] = make_pair(y, x);
    start[i] = 1;
  }
  int s = n;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d", &t);
    if (t == 3) {
      Q[i] = i;
    }
    if (t == 1) {
      start[++s] = i;
      scanf("%d %d", &x, &y);
      a[s] = make_pair(y, x);
    }
    if (t == 2) {
      scanf("%d", &x);
      finish[x] = i;
    }
  }
  for (int i = 1; i <= s; i++)
    if (!finish[i]) finish[i] = q;
  for (int i = 1; i <= s; i++) {
    if (start[i]) {
      add(1, 1, q, start[i], finish[i], i);
    }
  }
  solve(1, 1, q);
  return 0;
}
