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;
template <typename T>
inline void setmin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
inline void setmax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const int MAX = (1 << 14);
const int INF = 1e9 + 7;
const long long BIG_INF = 1e18 + 5;
int POCZ[MAX], KON[MAX];
long long MOD[MAX];
long long n, k, q;
pair<vector<int>, vector<int> > tree[2 * MAX];
vector<pair<int, int> > tab;
void dodaj(int pocz, int kon, int value) {
  pocz += MAX, kon += MAX;
  tree[pocz].first.push_back(value);
  if (pocz != kon) {
    tree[kon].first.push_back(value);
  }
  while (pocz / 2 != kon / 2) {
    if (!(pocz & 1)) tree[pocz + 1].first.push_back(value);
    if (kon & 1) tree[kon - 1].first.push_back(value);
    pocz /= 2;
    kon /= 2;
  }
}
vector<int> knapsack(vector<int> base, vector<int> &nowe) {
  for (auto u : nowe) {
    int koszt = tab[u].first;
    int waga = tab[u].second;
    for (int i = int((base).size()) - 1; i >= waga; i--)
      base[i] = max(base[i], base[i - waga] + koszt);
  }
  return base;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  MOD[0] = 1;
  for (int i = 1; i < MAX; i++)
    MOD[i] = (MOD[i - 1] * (long long)(1e7 + 19)) % INF;
  fill(KON, KON + MAX, MAX - 1);
  fill(POCZ, POCZ + MAX, MAX);
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    tab.push_back({a, b});
    POCZ[i] = 0;
  }
  cin >> q;
  int ile_q1 = 0;
  int ile_q3 = 0;
  vector<int> zapytania;
  int pom = 1;
  bool czy_zmiana = 1;
  for (int i = 0; i < q; i++) {
    int znak;
    cin >> znak;
    if (znak == 1) {
      czy_zmiana = 1;
      int a, b;
      cin >> a >> b;
      POCZ[n + ile_q1] = ile_q3;
      tab.push_back({a, b});
      ile_q1++;
    } else if (znak == 2) {
      czy_zmiana = 1;
      int a;
      cin >> a;
      a--;
      KON[a] = ile_q3 - 1;
    } else {
      if (czy_zmiana) {
        ile_q3++;
        zapytania.push_back(pom);
        pom = 1;
        czy_zmiana = 0;
      } else
        pom++;
    }
  }
  zapytania.push_back(pom);
  for (int i = 0; i < n + ile_q1; i++) {
    if (POCZ[i] > KON[i]) continue;
    dodaj(POCZ[i], KON[i], i);
  }
  vector<int> temp;
  temp.resize(k + 1);
  tree[1].second = temp;
  for (int i = 2; i < 2 * MAX; i++)
    tree[i].second = knapsack(tree[i / 2].second, tree[i].first);
  for (int i = MAX; i < MAX + ile_q3; i++) {
    long long odp = 0;
    for (int j = 1; j <= k; j++)
      odp = (odp + tree[i].second[j] * MOD[j - 1]) % INF;
    cout << odp << '\n';
    for (int j = 1; j < zapytania[i - MAX + 1]; j++) cout << odp << '\n';
  }
}
