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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
struct TNode {
  int L, R;
  int left, right;
  vector<int> toAdd;
  TNode(int L = -1, int R = -1) : L(L), R(R), left(-1), right(-1), toAdd() {}
};
const int MAXT = 110000;
TNode t[MAXT];
int tc = 0;
int buildTree(int L, int R) {
  int node = tc++;
  t[node] = TNode(L, R);
  if (R - L > 1) {
    int M = (L + R) / 2;
    t[node].left = buildTree(L, M);
    t[node].right = buildTree(M, R);
  }
  return node;
}
void addQ(int node, int L, int R, int x) {
  TNode &n = t[node];
  if (max(L, n.L) >= min(R, n.R)) return;
  if (L <= n.L && n.R <= R) {
    n.toAdd.push_back(x);
    return;
  }
  addQ(n.left, L, R, x);
  addQ(n.right, L, R, x);
}
const int maxn = 20000;
int act[maxn], w[maxn], v[maxn];
const int maxq = 31000;
int type[maxq], id[maxq], ansq[maxq];
const int maxk = 1001;
int dp[maxk];
const long long P = 1000000000 + 7;
const long long Q = 10000000 + 19;
vector<pair<int, int> > ops;
void change(int i, int x) {
  ops.push_back(make_pair(i, dp[i]));
  dp[i] = x;
}
void revert() {
  int i = ops.back().first, x = ops.back().second;
  dp[i] = x;
  ops.pop_back();
}
int n, k, q;
void traverse(int node) {
  TNode &n = t[node];
  int sz = ops.size();
  for (int id : n.toAdd) {
    for (int i = k - w[id]; i >= 0; --i) {
      if (dp[i] < 0) continue;
      if (dp[i + w[id]] < dp[i] + v[id]) change(i + w[id], dp[i] + v[id]);
    }
  }
  if (n.R - n.L == 1) {
    if (type[n.L] == 3) {
      long long ans = 0, t = 1;
      long long x = 0;
      for (int i = 1; i <= (int)(k); ++i) {
        uax(x, (long long)dp[i]);
        ans += t * x;
        ans %= P;
        t *= Q;
        t %= P;
      }
      ansq[n.L] = ans;
    }
  } else {
    traverse(n.left);
    traverse(n.right);
  }
  while (ops.size() > sz) revert();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  cin >> n >> k;
  for (int i = 0; i < (int)(n); ++i) {
    act[i] = 0;
    cin >> v[i] >> w[i];
  }
  cin >> q;
  int root = buildTree(0, q);
  for (int i = 0; i < (int)(q); ++i) {
    cin >> type[i];
    if (type[i] == 1) {
      cin >> v[n] >> w[n];
      act[n] = i;
      id[i] = n++;
    } else if (type[i] == 2) {
      cin >> id[i];
      --id[i];
      addQ(root, act[id[i]], i, id[i]);
      act[id[i]] = -1;
    }
  }
  for (int i = 0; i < (int)(n); ++i)
    if (act[i] != -1) addQ(root, act[i], q, i);
  traverse(root);
  for (int i = 0; i < (int)(q); ++i) {
    if (type[i] == 3) cout << ansq[i] << '\n';
  }
  return 0;
}