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 MAXSIZE = 30000020;
const int base = 10000019, mod = 1000000007;
int bufpos;
char buf[MAXSIZE];
void init() {
  buf[fread(buf, 1, MAXSIZE, stdin)] = '\0';
  bufpos = 0;
}
int readint() {
  int val = 0;
  for (; !isdigit(buf[bufpos]); bufpos++)
    ;
  for (; isdigit(buf[bufpos]); bufpos++) val = val * 10 + buf[bufpos] - '0';
  return val;
}
char readchar() {
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  return buf[bufpos++];
}
int readstr(char *s) {
  int cur = 0;
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  for (; !isspace(buf[bufpos]); bufpos++) s[cur++] = buf[bufpos];
  s[cur] = '\0';
  return cur;
}
const int maxn = 15002;
const int maxq = 30002;
int v[maxn], w[maxn];
inline void tense(int &x, int y) {
  if (x < y) x = y;
}
int k;
int lg[maxq * 4];
struct segtree {
  vector<int> a[maxq * 4];
  int ul, ur, id;
  int n;
  void update(int o, int l, int r) {
    if (ul <= l && ur >= r) {
      a[o].push_back(id);
      return;
    }
    int mid = (l + r) / 2;
    if (ul <= mid) update(o * 2, l, mid);
    if (ur > mid) update(o * 2 + 1, mid + 1, r);
  }
  void add(int l, int r, int p) {
    ul = l, ur = r, id = p;
    if (l > r) return;
    update(1, 1, n);
  }
  int dp[20][1002];
  void dfs(int o, int l, int r) {
    memcpy(dp[lg[o]], dp[lg[o] - 1], (k + 1) * 4);
    int *qwq = dp[lg[o]];
    for (auto it : a[o]) {
      for (int i = k; i >= w[it]; i--) tense(qwq[i], qwq[i - w[it]] + v[it]);
    }
    if (l == r) {
      long long res = 0;
      for (int i = k; i; i--) {
        res = (res * base + qwq[i]) % mod;
      }
      printf("%lld\n", res);
      return;
    }
    int mid = (l + r) / 2;
    dfs(o * 2, l, mid);
    dfs(o * 2 + 1, mid + 1, r);
  }
} t;
int st[maxn], en[maxn];
int main() {
  init();
  int n = readint();
  k = readint();
  for (int i = 1; i < maxq * 4; i++) lg[i] = lg[i / 2] + 1;
  int now = 0;
  memset(en, -1, sizeof(en));
  for (int i = 1; i <= n; i++) v[i] = readint(), w[i] = readint(), st[i] = 1;
  int q = readint();
  while (q--) {
    int op = readint();
    if (op == 1)
      v[++n] = readint(), w[n] = readint(), st[n] = now + 1;
    else if (op == 2)
      en[readint()] = now;
    else
      now++;
  }
  t.n = now;
  for (int i = 1; i <= n; i++) t.add(st[i], en[i] == -1 ? now : en[i], i);
  t.dfs(1, 1, now);
}
