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 long long mod = 1000000007, P = 10000000 + 19;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int N = 30100;
int dp[30][N];
int n, k, st[N], ed[N], ty, it, qur[N], q;
pair<int, int> p[N];
vector<int> ps;
void query(int d, int l, int r, vector<int> item) {
  for (int i = 0; i < k + 1; i++) dp[d][i] = dp[d - 1][i];
  for (int j = 0; j < ((int)(item).size()); j++)
    if (st[item[j]] <= l && ed[item[j]] >= r) {
      int v = p[item[j]].first, w = p[item[j]].second;
      for (int i = k + 1 - 1; i >= 0; i--) {
        if (i >= w) dp[d][i] = max(dp[d][i], dp[d][i - w] + v);
      }
    }
  if (l == r) {
    long long ret = 0;
    for (int i = k + 1 - 1; i >= 1; i--) ret = (ret * P + dp[d][i]) % mod;
    for (int j = 0; j < qur[l]; j++) printf("%d\n", (int)ret);
  } else {
    int md = (l + r) >> 1;
    vector<int> lt, rt;
    for (int j = 0; j < ((int)(item).size()); j++) {
      if (st[item[j]] <= l && ed[item[j]] >= r) continue;
      if (ed[item[j]] <= md)
        lt.push_back(item[j]);
      else if (st[item[j]] > md)
        rt.push_back(item[j]);
      else
        lt.push_back(item[j]), rt.push_back(item[j]);
    }
    query(d + 1, l, md, lt);
    query(d + 1, md + 1, r, rt);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n + 1; i++)
    scanf("%d%d", &p[i].first, &p[i].second), st[i] = 1;
  scanf("%d", &q);
  int cur = 1, im = n;
  for (int i = 0; i < q; i++) {
    scanf("%d", &ty);
    if (ty == 3) {
      qur[cur]++;
    } else if (ty == 2) {
      scanf("%d", &it);
      ed[it] = cur++;
    } else {
      im++;
      scanf("%d%d", &p[im].first, &p[im].second);
      st[im] = ++cur;
    }
  }
  for (int i = 1; i < im + 1; i++) ps.push_back(i);
  for (int i = 1; i < im + 1; i++)
    if (ed[i] == 0) ed[i] = cur;
  query(1, 1, cur, ps);
}
