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 = 1e5;
const int p = 1e7 + 19;
const int mod = 1e9 + 7;
int n, a[N], st[N], en[N], q, k, timer, m, v[N], w[N], ans[N];
vector<int> T[N];
void upd(int x, int l, int r, int v = 1, int tl = 1, int tr = timer) {
  if (tl > r || tr < l || tl > tr || l > r) return;
  if (l <= tl && tr <= r) {
    T[v].push_back(x);
    return;
  }
  int tm = (tl + tr) / 2;
  upd(x, l, r, v * 2, tl, tm), upd(x, l, r, v * 2 + 1, tm + 1, tr);
}
void insert(vector<int> &dp, int x) {
  vector<int> ans = dp;
  for (int i = w[x]; i <= k; i++) ans[i] = max(dp[i - w[x]] + v[x], ans[i]);
  dp = ans;
}
void go(int v = 1, int tl = 1, int tr = timer,
        vector<int> dp = vector<int>(k + 1)) {
  for (auto it : T[v]) insert(dp, it);
  if (tl == tr) {
    ans[tl] = 0;
    int pw = 1;
    int d = 0;
    for (int i = 1; i <= k; i++) {
      d = max(d, dp[i]);
      ans[tl] = (ans[tl] + (1ll * d * pw) % mod) % mod;
      pw = (1ll * pw * p) % mod;
    }
    return;
  }
  int tm = (tl + tr) / 2;
  go(v * 2, tl, tm, dp), go(v * 2 + 1, tm + 1, tr, dp);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    ++m;
    scanf("%d%d", &v[m], &w[m]);
  }
  scanf("%d", &q);
  memset(st, -1, sizeof st);
  memset(en, -1, sizeof en);
  for (int i = 1, type, id; i <= q; i++) {
    scanf("%d", &type);
    if (type == 3) timer++;
    if (type == 2) {
      scanf("%d", &id);
      en[id] = timer;
    }
    if (type == 1) {
      ++m;
      scanf("%d%d", &v[m], &w[m]);
      st[m] = timer + 1;
    }
  }
  for (int i = 1; i <= m; i++)
    upd(i, (st[i] == -1 ? 1 : st[i]), (en[i] == -1 ? timer : en[i]));
  go();
  for (int i = 1; i <= timer; i++) printf("%d\n", ans[i]);
  return 0;
}
