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;
int W;
int timer = 0;
int value[(1 << 15)], mass[(1 << 15)], tin[(1 << 15)], tout[(1 << 15)];
vector<int> items[2 * (1 << 15)];
int dp[20][1010];
int ans[(1 << 15)];
void update(int level, int v, int w) {
  for (int i = W; i >= w; i--)
    dp[level][i] = max(dp[level][i], dp[level][i - w] + v);
}
void solve(int level, int id, int low = 0, int high = (1 << 15)) {
  int i;
  for ((i) = 0; (i) < (int)(items[id].size()); (i)++)
    update(level, value[items[id][i]], mass[items[id][i]]);
  if (level == 15) {
    long long sum = 0;
    for (i = W; i >= 1; i--)
      sum = (sum * 10000019 + dp[level][i]) % 1000000007ll;
    ans[id - (1 << 15)] = sum;
    return;
  }
  int mid = (low + high) / 2;
  for ((i) = 0; (i) < (int)(W + 1); (i)++) dp[level + 1][i] = dp[level][i];
  solve(level + 1, 2 * id, low, mid);
  for ((i) = 0; (i) < (int)(W + 1); (i)++) dp[level + 1][i] = dp[level][i];
  solve(level + 1, 2 * id + 1, mid, high);
}
void add_query(int item_number, int from, int to, int id = 1, int low = 0,
               int high = (1 << 15)) {
  if (from <= low && to >= high) {
    items[id].push_back(item_number);
    return;
  }
  int mid = (low + high) / 2;
  if (from < mid) add_query(item_number, from, to, 2 * id, low, mid);
  if (to > mid) add_query(item_number, from, to, 2 * id + 1, mid, high);
}
int main(void) {
  int N, Q, i;
  for ((i) = 0; (i) < (int)((1 << 15)); (i)++)
    value[i] = mass[i] = tin[i] = tout[i] = -1;
  cin >> N >> W;
  for ((i) = 0; (i) < (int)(N); (i)++) {
    scanf("%d%d", &value[i], &mass[i]);
    tin[i] = 0;
  }
  cin >> Q;
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &value[N], &mass[N]);
      tin[N] = timer;
      N++;
    } else if (type == 2) {
      int x;
      scanf("%d", &x);
      x--;
      tout[x] = timer;
    } else {
      timer++;
    }
  }
  for ((i) = 0; (i) < (int)(N); (i)++)
    if (tout[i] == -1) tout[i] = timer;
  for ((i) = 0; (i) < (int)(N); (i)++)
    if (tin[i] != tout[i]) add_query(i, tin[i], tout[i]);
  solve(0, 1);
  for ((i) = 0; (i) < (int)(timer); (i)++) printf("%d\n", ans[i]);
  return 0;
}
