Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
template <class T>
inline bool getmax(T &a, const T &b) {
  return a < b ? (a = b, true) : false;
}
struct Item {
  int V, W, L, R;
  Item() {}
  Item(const int _V, const int _W, const int _L)
      : V(_V), W(_W), L(_L), R(35000) {}
};
int N, K;
long long ANS[30000];
void Solve(const int l, const int r, const vector<Item> &items,
           const vector<pair<int, int> > &queries, const long long *pre_dp) {
  long long *dp = new long long[K + 1];
  for (int i = 0; i <= K; i++) dp[i] = pre_dp[i];
  const int mid = (l + r) / 2;
  vector<Item> left_items, rigt_items;
  for (const Item &item : items) {
    if (item.L <= l && r <= item.R) {
      for (int i = K - item.W; i >= 0; i--)
        getmax(dp[i + item.W], dp[i] + item.V);
    } else {
      if (item.L <= mid) left_items.push_back(item);
      if (mid + 1 <= item.R) rigt_items.push_back(item);
    }
  }
  if (l == r) {
    long long ans = 0, p = 1;
    for (int m = 1; m <= K; m++, (p *= 10000000 + 19) %= MOD)
      (ans += dp[m] * p) %= MOD;
    for (const auto &q : queries) assert(q.first == r), ANS[q.second] = ans;
    return;
  }
  vector<pair<int, int> > left_queries, rigt_queries;
  for (const auto &q : queries)
    (q.first <= mid ? left_queries : rigt_queries).push_back(q);
  Solve(l, mid, left_items, left_queries, dp);
  Solve(mid + 1, r, rigt_items, rigt_queries, dp);
  delete[] dp;
}
int main() {
  while (scanf("%d%d", &N, &K) == 2) {
    vector<Item> item;
    int tick = 0;
    for (int i = 0, v, w; i < N; i++)
      scanf("%d%d", &v, &w), item.push_back(Item(v, w, tick++));
    int querycount;
    scanf("%d", &querycount);
    vector<pair<int, int> > queries;
    int ans_length = 0;
    for (int i = 0, type; i < querycount; i++) {
      scanf("%d", &type);
      if (type == 1) {
        int v, w;
        scanf("%d%d", &v, &w);
        item.push_back(Item(v, w, tick++));
      } else if (type == 2) {
        int x;
        scanf("%d", &x), --x;
        item[x].R = (tick++) - 1;
      } else if (type == 3) {
        queries.push_back(make_pair(tick - 1, ans_length++));
      } else
        assert(0);
    }
    long long *dp = new long long[K + 1];
    for (int i = 0; i <= K; i++) dp[i] = 0;
    Solve(0, tick - 1, item, queries, dp);
    delete[] dp;
    for (int i = 0; i < ans_length; i++) printf("%lld\n", ANS[i]);
  }
  return 0;
}