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>
void write(ostream& out, T t) {
  out << t << "\n";
}
template <typename T, typename... Args>
void write(ostream& out, T t, Args... args) {
  out << t << ' ';
  write(out, args...);
}
template <typename... Args>
void debug(Args... args) {}
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
const long long modn2 = 10000019;
inline long long mod(long long x) { return x % modn; }
const int MAXK = 1003, MAXQ = 2 * 30004 + 10010;
const int MAXN = MAXQ;
const long long INF = 9000000000LL;
int n, m, k;
vector<pair<long long, int> > s;
long long dp[15009][MAXK];
void prep_cin() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
struct qry {
  int t, id;
};
vector<qry> q;
int L[MAXN], R[MAXN];
int acu[MAXN];
long long pot[MAXK];
int h;
inline long long f(int h) {
  long long res = 0;
  long long mx = 0;
  for (int a = 1; a <= k; a++) {
    mx = mod(max(mx, dp[h][a]));
    res = mod(res + mod(mx * pot[a - 1]));
  }
  return res;
}
inline bool sup(int i, int j, int l, int r) { return (i < l && j > r); }
inline void add(int id) {
  h++;
  int pes = s[id].second;
  long long val = s[id].first;
  for (int a = 0; a <= k; a++) {
    dp[h][a] = dp[h - 1][a];
    if (a >= pes && dp[h - 1][a - pes] != -INF)
      dp[h][a] = max(dp[h][a], dp[h - 1][a - pes] + val);
  }
}
void go(int i, int j) {
  if (i == j) {
    if (q[i].t == 3) printf("%lld\n", f(h));
    return;
  }
  int m = (i + j) / 2;
  if (acu[m] - (i == 0 ? 0 : acu[i - 1]) != 0) {
    int old = h;
    for (int a = m + 1; a <= j; a++) {
      if (q[a].t == 1 || q[a].t == 3) continue;
      int id = q[a].id;
      if (sup(L[id], R[id], i, m)) add(id);
    }
    go(i, m);
    h = old;
  }
  if (acu[j] - acu[m] != 0) {
    int old = h;
    for (int a = i; a <= m; a++) {
      if (q[a].t == 2 || q[a].t == 3) continue;
      int id = q[a].id;
      if (sup(L[id], R[id], m + 1, j)) add(id);
    }
    go(m + 1, j);
    h = old;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  pot[0] = 1;
  for (int a = 1; a < k; a++) pot[a] = mod(pot[a - 1] * modn2);
  const int inf = 1e7;
  for (int a = 0; a < n; a++) {
    pair<long long, int> x;
    scanf("%lld %d", &x.first, &x.second);
    s.push_back(x);
    q.push_back({1, a});
    L[a] = a;
    R[a] = inf;
  }
  scanf("%d", &m);
  for (int a = 0; a < m; a++) {
    qry p;
    scanf("%d", &p.t);
    if (q.size() > 0) acu[q.size()] = acu[q.size() - 1];
    if (p.t == 1) {
      pair<long long, int> x;
      scanf("%lld %d", &x.first, &x.second);
      s.push_back(x);
      p.id = (int)s.size() - 1;
      L[p.id] = q.size();
      R[p.id] = inf;
    } else if (p.t == 2) {
      scanf("%d", &p.id);
      p.id--;
      R[p.id] = q.size();
    } else
      acu[q.size()]++;
    q.push_back(p);
  }
  n = s.size();
  m = q.size();
  for (int a = 0; a <= k; a++) dp[0][a] = -INF;
  dp[0][0] = 0;
  go(0, m - 1);
}
