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;
const int mod = 1000 * 1000 * 1000 + 7;
const int INF = 1e9 + 100;
const long long LINF = 1e18 + 100;
const int prm = 1e7 + 19;
int n, k;
int val[15050], wei[15050];
int q;
int op[30100];
int beg[15050];
vector<int> seg[30100 * 4];
long long dp[22][1050];
inline void additem(int x, long long *dp) {
  for (int i = 1050 - 1; i >= wei[x]; i--)
    dp[i] = max(dp[i], dp[i - wei[x]] + val[x]);
}
void add(int L, int R, int x, int v = 1, int l = 0, int r = 30100 - 1) {
  if (l > R || r < L) return;
  if (l >= L && r <= R) {
    seg[v].push_back(x);
    return;
  }
  int mid = (r + l) / 2;
  add(L, R, x, 2 * v, l, mid);
  add(L, R, x, 2 * v + 1, mid + 1, r);
}
void solve(int v = 1, int l = 0, int r = 30100 - 1, int lvl = 1) {
  memcpy(dp[lvl], dp[lvl - 1], sizeof(dp[lvl]));
  for (int u : seg[v]) additem(u, dp[lvl]);
  if (l == r) {
    if (op[l] == 3) {
      long long ans = 0;
      for (int i = k; i >= 1; i--) ans = (ans * prm + dp[lvl][i]) % mod;
      cout << ans << '\n';
    }
  } else {
    int mid = (r + l) / 2;
    solve(2 * v, l, mid, lvl + 1);
    solve(2 * v + 1, mid + 1, r, lvl + 1);
  }
}
int32_t main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = (1); i < (int)(n + 1); i++) cin >> val[i] >> wei[i], beg[i] = 1;
  cin >> q;
  for (int i = (1); i < (int)(q + 1); i++) {
    cin >> op[i];
    if (op[i] == 1) {
      n++;
      cin >> val[n] >> wei[n];
      beg[n] = i;
    } else if (op[i] == 2) {
      int x;
      cin >> x;
      add(beg[x], i, x);
      beg[x] = -1;
    }
  }
  for (int i = (1); i < (int)(n + 1); i++) {
    if (beg[i] >= 0) add(beg[i], 30100 - 1, i);
  }
  solve();
  return 0;
}
