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;
vector<int> seg[(1 << 15) << 1];
int st[(1 << 15)], m[(1 << 15)], v[(1 << 15)], k;
bool in[(1 << 15)];
bool ask[(1 << 15)];
const int Q = 1e9 + 7, P = 1e7 + 19;
void add(int u, int l, int r, int ql, int qr, int x) {
  if (ql == qr) return;
  if (l == ql && r == qr) {
    seg[u].push_back(x);
  } else {
    int m = (l + r) / 2;
    if (qr <= m)
      add(u << 1, l, m, ql, qr, x);
    else if (ql >= m)
      add(u << 1 | 1, m, r, ql, qr, x);
    else {
      add(u << 1, l, m, ql, m, x);
      add(u << 1 | 1, m, r, m, qr, x);
    }
  }
}
void solve(int u, int l, int r, int up, int pre[]) {
  if (l >= up) return;
  int now[1010];
  memcpy(now, pre, sizeof(now));
  for (int i = 0; i < seg[u].size(); i++) {
    int x = seg[u][i];
    for (int j = k - m[x]; j >= 0; j--) {
      if (now[j] < 0)
        continue;
      else if (now[j] + v[x] > now[j + m[x]])
        now[j + m[x]] = now[j] + v[x];
    }
  }
  if (l + 1 == r) {
    int ans = 0, tmp = 0, mul = 1;
    for (int i = 1; i <= k; i++) {
      if (now[i] > tmp) tmp = now[i];
      ans = (ans + 1LL * tmp * mul) % Q;
      mul = 1LL * mul * P % Q;
    }
    printf("%d\n", ans);
  } else {
    solve(u << 1, l, (l + r) / 2, up, now);
    solve(u << 1 | 1, (l + r) / 2, r, up, now);
  }
}
int main() {
  int n, cnt, q, now = 0;
  scanf("%d%d", &n, &k);
  cnt = n + 1;
  for (int i = 1; i <= n; i++)
    st[i] = 0, in[i] = 1, scanf("%d%d", &v[i], &m[i]);
  scanf("%d", &q);
  while (q--) {
    int tp, x, y;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d%d", &v[cnt], &m[cnt]);
      in[cnt] = 1;
      st[cnt++] = now;
    } else if (tp == 2) {
      scanf("%d", &x);
      in[x] = 0;
      add(1, 0, (1 << 15), st[x], now, x);
    } else {
      now++;
    }
  }
  for (int i = 1; i < cnt; i++) {
    if (in[i]) add(1, 0, (1 << 15), st[i], now, i);
  }
  int pre[1010];
  memset(pre, -1, sizeof(pre));
  pre[0] = 0;
  solve(1, 0, (1 << 15), now, pre);
  return 0;
}
