Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long rt = 1e7 + 19;
const long long mod = 1e9 + 7;
const int inf = 2e9;
int n, k, qry;
vector<int> obj[222222];
int sack[22][11111], v[44444], w[44444], go[44444], com[44444];
bool respos[44444];
void add(int l, int r, int ql, int qr, int cur, int object) {
  if (ql <= l and r <= qr) {
    obj[cur].push_back(object);
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) {
    add(l, mid, ql, qr, cur + cur, object);
  }
  if (mid < qr) {
    add(mid + 1, r, ql, qr, cur + cur + 1, object);
  }
}
void dfs(int l, int r, int cur, int level) {
  for (int i = 0; i <= k; i++) {
    sack[level][i] = sack[level - 1][i];
  }
  for (auto u : obj[cur]) {
    for (int i = k; i >= 0; i--) {
      if (i + w[u] > k or sack[level][i] == -inf) continue;
      sack[level][i + w[u]] = max(sack[level][i + w[u]], sack[level][i] + v[u]);
    }
  }
  if (l == r) {
    if (respos[l] == false) return;
    long long res = 0, mx = 0, p = 1;
    for (int m = 1; m <= k; m++) {
      mx = max(mx, 1LL * sack[level][m]);
      res = (p * mx + res) % mod;
      p = (p * rt) % mod;
    }
    printf("%lld\n", res);
    return;
  }
  int mid = (l + r) >> 1;
  dfs(l, mid, cur + cur, level + 1);
  dfs(mid + 1, r, cur + cur + 1, level + 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", v + i, w + i);
    com[i] = 1;
  }
  scanf("%d", &qry);
  for (int i = 1; i <= qry; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      n++;
      scanf("%d%d", v + n, w + n);
      com[n] = i;
    } else if (type == 2) {
      int q;
      scanf("%d", &q);
      go[q] = i;
    } else {
      respos[i] = true;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (go[i] == 0) go[i] = qry;
    add(1, qry, com[i], go[i], 1, i);
  }
  fill(sack[0] + 1, sack[0] + 1111, -inf);
  dfs(1, qry, 1, 1);
  return 0;
}