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 N = 40005, K = 1005;
using ll = long long;
struct TNode {
  int l, r;
  vector<pair<int, int>> v;
} t[N << 2];
inline int lc(int pos) { return pos << 1; }
inline int rc(int pos) { return pos << 1 | 1; }
void build(int pos, int l, int r) {
  t[pos].l = l;
  t[pos].r = r;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(lc(pos), l, mid);
  build(rc(pos), mid + 1, r);
}
void modify(int pos, int l, int r, pair<int, int> v) {
  if (t[pos].l == l && t[pos].r == r) {
    t[pos].v.push_back(v);
    return;
  }
  int mid = (t[pos].l + t[pos].r) >> 1;
  if (r <= mid)
    modify(lc(pos), l, r, v);
  else if (l > mid)
    modify(rc(pos), l, r, v);
  else
    modify(lc(pos), l, mid, v), modify(rc(pos), mid + 1, r, v);
}
int n, kk, q, idx = 1;
struct Query {
  int v, w, l, r;
} e[N];
int dp[K];
const int BASE = 10000019, P = 1000000007;
void solve(int pos, int l, int r, int dp[]) {
  for (auto &i : t[pos].v)
    for (int j = kk; j >= i.second; --j)
      dp[j] = max(dp[j], dp[j - i.second] + i.first);
  if (l == r) {
    ll ans = 0;
    for (int i = kk; i > 0; --i) ans = (ans * BASE + dp[i]) % P;
    cout << ans << endl;
    return;
  }
  int mid = (l + r) >> 1;
  int f[K];
  memcpy(f, dp, sizeof(f));
  solve(lc(pos), l, mid, f);
  memcpy(f, dp, sizeof(f));
  solve(rc(pos), mid + 1, r, f);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> kk;
  int t1, t2;
  for (int i = 1; i <= n; i++) {
    cin >> t1 >> t2;
    e[i] = {t1, t2, idx, -1};
  }
  cin >> q;
  while (q--) {
    int opt;
    cin >> opt;
    if (opt == 1) {
      cin >> t1 >> t2;
      e[++n] = {t1, t2, idx, -1};
    } else if (opt == 2)
      cin >> t1, e[t1].r = idx - 1;
    else
      ++idx;
  }
  --idx;
  build(1, 1, idx);
  for (int i = 1; i <= n; i++) {
    if (e[i].r == -1) e[i].r = idx;
    if (e[i].l <= e[i].r) modify(1, e[i].l, e[i].r, {e[i].v, e[i].w});
  }
  solve(1, 1, idx, dp);
  return 0;
}
