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 maxn = 3e4 + 5, maxk = 1e3 + 5, maxlog = 20, P = 1e7 + 19,
          md = 1e9 + 7;
int n, k, w[maxn], v[maxn], st[maxn], ft[maxn], q;
long long dp[maxlog][maxk], pw[maxk];
vector<int> seg[maxn << 2];
bool ask[maxn << 2];
void add(int ql, int qr, int val, int l = 0, int r = maxn, int id = 1) {
  if (max(ql, l) >= min(qr, r)) return;
  if (ql <= l && qr >= r) {
    seg[id].push_back(val);
    return;
  }
  int mid = l + r >> 1;
  add(ql, qr, val, l, mid, id << 1);
  add(ql, qr, val, mid, r, id << 1 | 1);
}
void dfs(int l = 0, int r = maxn, int id = 1, int h = 0) {
  for (auto x : seg[id])
    for (int i = maxk - 1; i >= w[x]; i--)
      dp[h][i] = max(dp[h][i], dp[h][i - w[x]] + v[x]);
  if (r - l == 1 && ask[l]) {
    long long ans = 0;
    for (int i = 1; i <= k; i++) ans = (ans + dp[h][i] * pw[i - 1] % md) % md;
    cout << ans << '\n';
  }
  if (r - l == 1) return;
  int mid = l + r >> 1;
  for (int i = 0; i < maxk; i++) dp[h + 1][i] = dp[h][i];
  dfs(l, mid, id << 1, h + 1);
  for (int i = 0; i < maxk; i++) dp[h + 1][i] = dp[h][i];
  dfs(mid, r, id << 1 | 1, h + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pw[0] = 1;
  for (int i = 1; i < maxk; i++) pw[i] = pw[i - 1] * P % md;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
    st[i] = 0, ft[i] = maxn;
  }
  cin >> q;
  for (int ti = 1; ti <= q; ti++) {
    int type;
    cin >> type;
    if (type == 1) {
      cin >> v[++n] >> w[n];
      st[n] = ti;
      ft[n] = maxn;
    } else if (type == 2) {
      int x;
      cin >> x;
      ft[x] = ti;
    } else
      ask[ti] = 1;
  }
  for (int i = 1; i <= n; i++) add(st[i], ft[i], i);
  dfs();
}
