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 int inf = 1e9 + 333;
const long long linf = 1e18 + inf;
const int N = 35000 + 5;
const int K = 1000 + 5;
int n, k, q;
int a[N], b[N], st[N], nd[N], tme[N], ans[N], dp[K];
vector<pair<int, int> > t[N << 2];
void add(int x, int l, int r, int x1, int x2, int a, int b) {
  if (x2 < l or r < x1) return;
  if (x1 <= l and r <= x2) {
    t[x].push_back(pair<int, int>(a, b));
    return;
  }
  int m = (l + r) >> 1;
  add(x + x, l, m, x1, x2, a, b);
  add(x + x + 1, m + 1, r, x1, x2, a, b);
}
void solve(int x, int l, int r) {
  vector<int> backup;
  for (int i = 0; i <= k; i++) {
    backup.push_back(dp[i]);
  }
  for (__typeof((t[x]).begin()) it = (t[x]).begin(); it != (t[x]).end(); it++) {
    int a = it->first;
    int b = it->second;
    for (int i = k; i >= b; i--) {
      dp[i] = max(dp[i], dp[i - b] + a);
    }
  }
  if (l == r) {
    for (int i = k; i >= 1; i--) {
      ans[l] = ((long long)ans[l] * (int)(1e7 + 19) + dp[i]) % (int)(1e9 + 7);
    }
  } else {
    int m = (l + r) >> 1;
    solve(x + x, l, m);
    solve(x + x + 1, m + 1, r);
  }
  for (int i = 0; i <= k; i++) dp[i] = backup[i];
  backup.clear();
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", a + i, b + i);
    st[i] = 1;
    nd[i] = N - 1;
  }
  scanf("%d", &q);
  vector<int> v;
  int cnt = n;
  for (int i = 1; i <= q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      ++cnt;
      scanf("%d %d", a + cnt, b + cnt);
      st[cnt] = i + n;
      nd[cnt] = N - 1;
    } else if (type == 2) {
      int x;
      scanf("%d", &x);
      nd[x] = i - 1 + n;
    } else {
      v.push_back(i + n);
    }
  }
  for (int i = 1; i <= cnt; i++) {
    add(1, 1, N - 1, st[i], nd[i], a[i], b[i]);
  }
  solve(1, 1, N - 1);
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
    int x = *it;
    printf("%d\n", ans[x]);
  }
  return 0;
}
