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;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long fpow(long long n, long long k, int p = (int)1e9 + 7) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = 2 * acos((long double)0);
const int maxn = 15010;
const int maxk = 1010;
int n, k;
int v[maxn];
int w[maxn];
int l[maxn];
int r[maxn];
int f[maxk];
int p[maxk];
vector<pair<int, int> > query;
vector<pair<int, pair<int, int> > > itv;
priority_queue<pair<int, int> > pq;
int compare(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return make_pair(make_pair(-a.second.first, a.second.second), a.first) <
         make_pair(make_pair(-b.second.first, b.second.second), b.first);
}
void add(int v, int w, vector<pair<int, int> >& his) {
  for (int i = (k)-1; i >= (0); i--)
    if (i + w <= k && f[i + w] < f[i] + v) {
      his.push_back(make_pair(i + w, f[i + w]));
      f[i + w] = f[i] + v;
    }
}
void rollback(vector<pair<int, int> >& his, vector<pair<int, int> >& his2) {
  for (int i = (int((his).size())) - 1; i >= (0); i--) {
    int u = his[i].first;
    int v = his[i].second;
    f[u] = v;
  }
  for (int i = (0); i < (int((his2).size())); i++) {
    pq.push(his2[i]);
  }
}
void go(int L, int R) {
  while (int((itv).size()) && itv.back().second.first <= L) {
    pq.push(make_pair(itv.back().second.second, itv.back().first));
    itv.pop_back();
  }
  vector<pair<int, int> > his, his2;
  while (int((pq).size()) && pq.top().first >= R) {
    int u = pq.top().second;
    add(v[u], w[u], his);
    his2.push_back(pq.top());
    pq.pop();
  }
  if (L == R) {
    if (query[L].first == 3) {
      int ans = 0, dmax = 0;
      for (int i = (1); i < (k + 1); i++) {
        dmax = max(dmax, f[i]);
        ans = (ans + (long long)dmax * p[i - 1]) % MOD;
      }
      printf("%d\n", ans);
    }
  } else {
    go(L, (L + R) >> 1);
    go(((L + R) >> 1) + 1, R);
  }
  rollback(his, his2);
}
void solve() {
  p[0] = 1;
  for (int i = (1); i < (maxk); i++) p[i] = 10000019LL * p[i - 1] % MOD;
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n); i++) {
    scanf("%d%d", v + i, w + i);
    l[i] = i;
    r[i] = INF;
    query.push_back(make_pair(1, i));
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      v[n] = x;
      w[n] = y;
      l[n] = int((query).size());
      r[n] = INF;
      query.push_back(make_pair(op, n));
      n++;
    } else if (op == 2) {
      int x;
      scanf("%d", &x);
      x--;
      r[x] = int((query).size()) - 1;
      query.push_back(make_pair(op, x));
    } else if (op == 3) {
      query.push_back(make_pair(op, 0));
    }
  }
  for (int i = (0); i < (n); i++)
    itv.push_back(make_pair(i, make_pair(l[i], r[i])));
  sort((itv).begin(), (itv).end(), compare);
  go(0, int((query).size()) - 1);
}
int main() {
  solve();
  return 0;
}