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;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 3e4 + 10;
const int xm = 1e3 + 10;
const int sq = 320;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
const int mod = 1e9 + 7;
const int base = 1e7 + 19;
int n, k, q, ans, pw[xm], dp[xm], pd[17][xm];
pair<int, int> a[xn], b[xn];
bool flag[xn];
vector<int> seg[xn << 2];
void update(int id, int l, int r, int ql, int qr, int x) {
  if (qr <= l || r <= ql || qr <= ql) return;
  if (ql <= l && r <= qr) {
    seg[id].push_back(x);
    return;
  }
  int mid = l + r >> 1;
  update(id << 1, l, mid, ql, qr, x);
  update(id << 1 | 1, mid, r, ql, qr, x);
}
void get(int id, int l, int r, int h = 0) {
  for (int i = 0; i <= k; ++i) pd[h][i] = dp[i];
  for (int ind : seg[id])
    for (int i = k; i >= a[ind].second; --i)
      dp[i] = max(dp[i], dp[i - a[ind].second] + a[ind].first);
  if (r - l == 1 && flag[l]) {
    ans = 0;
    for (int i = 1; i <= k; ++i)
      ans = (ans + 1ll * pw[i - 1] * dp[i] % mod) % mod;
    cout << ans << '\n';
  }
  int mid = l + r >> 1;
  if (r - l > 1) get(id << 1, l, mid, h + 1), get(id << 1 | 1, mid, r, h + 1);
  for (int i = 0; i <= k; ++i) dp[i] = pd[h][i];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  pw[0] = 1;
  for (int i = 1; i < xm; ++i) pw[i] = 1ll * pw[i - 1] * base % mod;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i].first >> a[i].second;
  cin >> q;
  for (int i = 1; i <= n; ++i) b[i].second = q;
  for (int i = 1; i <= q; ++i) {
    int t, x;
    cin >> t;
    if (t == 1)
      ++n, cin >> a[n].first >> a[n].second, b[n] = {i, q};
    else if (t == 2)
      cin >> x, b[x].second = i;
    else
      flag[i] = true;
  }
  for (int i = 1; i <= n; ++i)
    update(1, 0, q + 1, b[i].first, b[i].second + 1, i);
  get(1, 0, q + 1);
  return 0;
}
