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 MAXN = 50005;
const int MAXK = 1005;
const int MAXQ = 50005;
const int P = 1e7 + 19;
const int Q = 1e9 + 7;
int n, k, q, exhibit_cnt, ask_cnt;
struct event {
  int type, x;
} evt[MAXQ];
int powp[MAXK];
int v[MAXQ], w[MAXQ];
int display_start[MAXQ], display_end[MAXQ];
vector<int> t[4 * MAXQ];
void update(int x, int l, int r, int ll, int rr, int v) {
  if (r < ll || l > rr || l > r) {
    return;
  }
  if (ll <= l && r <= rr) {
    return t[x].push_back(v);
  }
  int mid = l + r >> 1;
  update(2 * x, l, mid, ll, rr, v);
  update(2 * x + 1, mid + 1, r, ll, rr, v);
}
int dp[MAXN][MAXK];
int res[MAXQ];
void go(int x, int l, int r, int last) {
  for (int exhibit = last + 1; exhibit < last + 1 + (int)t[x].size();
       exhibit++) {
    for (int j = 0; j < MAXK; j++) {
      dp[exhibit][j] = 0;
    }
  }
  for (int exhibit = last + 1; exhibit < last + 1 + (int)t[x].size();
       exhibit++) {
    for (int mass = 1; mass <= k; mass++) {
      if (exhibit) dp[exhibit][mass] = dp[exhibit - 1][mass];
      if (mass >= w[t[x][exhibit - last - 1]]) {
        if (exhibit) {
          dp[exhibit][mass] =
              max(dp[exhibit][mass],
                  dp[exhibit - 1][mass - w[t[x][exhibit - last - 1]]] +
                      v[t[x][exhibit - last - 1]]);
        } else
          dp[exhibit][mass] =
              max(dp[exhibit][mass], v[t[x][exhibit - last - 1]]);
      }
    }
  }
  if (l == r) {
    for (int mass = 1; mass <= k; mass++) {
      res[l] += (1ll * dp[last + (int)t[x].size()][mass] * powp[mass - 1]) % Q;
      res[l] %= Q;
    }
    return;
  }
  int mid = l + r >> 1;
  go(2 * x, l, mid, last + (int)t[x].size());
  go(2 * x + 1, mid + 1, r, last + (int)t[x].size());
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < MAXQ; i++) {
    display_end[i] = 0x3f3f3f3f;
  }
  powp[0] = 1;
  for (int i = 1; i < MAXK; i++) {
    powp[i] = (1ll * powp[i - 1] * P) % Q;
  }
  for (int i = 1; i <= n; i++) {
    evt[i].type = 1;
    evt[i].x = ++exhibit_cnt;
    scanf("%d %d", v + evt[i].x, w + evt[i].x);
    display_start[i] = ask_cnt;
  }
  scanf("%d", &q);
  for (int i = n + 1; i <= n + q; i++) {
    scanf("%d", &evt[i].type);
    if (evt[i].type == 1) {
      evt[i].x = ++exhibit_cnt;
      scanf("%d %d", v + evt[i].x, w + evt[i].x);
      display_start[evt[i].x] = ask_cnt + 1;
    } else if (evt[i].type == 2) {
      scanf("%d", &evt[i].x);
      display_end[evt[i].x] = ask_cnt;
    } else {
      ask_cnt++;
    }
  }
  for (int i = 1; i <= exhibit_cnt; i++) {
    if (display_end[i] == 0) continue;
    if (display_start[i] == 0) display_start[i] = 1;
    if (display_start[i] > ask_cnt) continue;
    if (display_end[i] == 0x3f3f3f3f) display_end[i] = ask_cnt;
    update(1, 1, ask_cnt, display_start[i], display_end[i], i);
  }
  go(1, 1, ask_cnt, -1);
  for (int i = 1; i <= ask_cnt; i++) {
    printf("%d\n", res[i]);
  }
  return 0;
}
