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 = 30005;
const int P = 1e7 + 19;
const int mod = 1e9 + 7;
int n, k, dp[1005], Q, tot, qq[maxn << 2], an[maxn << 2], tt, po[maxn], ss, ans,
    now[maxn << 2];
vector<pair<int, int> > st[maxn << 2];
struct node {
  int st, en, v, w;
} d[maxn << 2];
vector<node> vec[maxn << 3];
inline void add(int &x, int y) { x + y >= mod ? x = x + y - mod : x += y; }
inline void change(int rt, int j, int x) {
  st[rt].push_back(make_pair(j, dp[j]));
  dp[j] = x;
}
inline void re(int rt) {
  int mx = 1061109567;
  for (int i = st[rt].size() - 1; i >= 0; --i) {
    int j = st[rt][i].first;
    dp[j] = st[rt][i].second;
    mx = min(mx, j);
  }
  for (int j = mx; j <= k; ++j) {
    add(ans, mod - 1LL * po[j - 1] * now[j] % mod);
    now[j] = max(dp[j], now[j - 1]);
    add(ans, 1LL * po[j - 1] * now[j] % mod);
  }
  st[rt].clear();
  vec[rt].clear();
}
void solve(int l, int r, int rt) {
  int mid = (l + r) >> 1;
  for (int i = 0; i < vec[rt].size(); ++i) {
    node x = vec[rt][i];
    if (x.st <= l && x.en >= r) {
      for (int j = k; j >= x.w; --j) {
        if (dp[j] < dp[j - x.w] + x.v) change(rt, j, dp[j - x.w] + x.v);
      }
      for (int j = x.w; j <= k; ++j) {
        add(ans, mod - 1LL * po[j - 1] * now[j] % mod);
        now[j] = max(dp[j], now[j - 1]);
        add(ans, 1LL * po[j - 1] * now[j] % mod);
      }
    } else {
      if (x.st <= mid) vec[rt << 1].push_back(x);
      if (x.en > mid) vec[rt << 1 | 1].push_back(x);
    }
  }
  if (l == r) {
    if (qq[l]) {
      an[++tt] = ans;
    }
    re(rt);
    return;
  }
  solve(l, mid, rt << 1);
  solve(mid + 1, r, rt << 1 | 1);
  re(rt);
}
int main() {
  scanf("%d%d", &n, &k);
  po[0] = 1;
  for (int i = 1; i <= k; ++i) po[i] = 1LL * po[i - 1] * P % mod;
  ss = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &d[i].v, &d[i].w);
    d[i].st = 1;
    d[i].en = 1061109567;
  }
  scanf("%d", &Q);
  tt = 0;
  for (int i = 1; i <= Q; ++i) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      d[++ss].st = i;
      d[ss].en = 1061109567;
      scanf("%d%d", &d[ss].v, &d[ss].w);
    } else if (opt == 2) {
      int x;
      scanf("%d", &x);
      d[x].en = i;
    } else
      qq[i] = 1;
  }
  ans = 0;
  for (int i = 1; i <= ss; ++i) vec[1].push_back(d[i]);
  solve(1, Q, 1);
  for (int i = 1; i <= tt; ++i) printf("%d\n", an[i]);
}
