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 = 4e4 + 7;
const int maxk = 1e3 + 7;
const int base = 1e7 + 19;
const int MOD = 1e9 + 7;
struct exhibits {
  int v, w, l, r;
} exb[maxn];
int k;
vector<pair<int, int> > seg[maxn << 2];
long long DP[maxk];
int qr();
void add(int p, int l, int r, int ql, int qr, int v, int w) {
  if (ql <= l && r <= qr) {
    seg[p].push_back(make_pair(v, w));
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) add(p << 1, l, mid, ql, qr, v, w);
  if (mid < qr) add(p << 1 | 1, mid + 1, r, ql, qr, v, w);
}
void solve(int p, int l, int r, long long* dp) {
  for (vector<pair<int, int> >::iterator pp = seg[p].begin();
       pp != seg[p].end(); pp++)
    for (int j = k; j >= pp->second; j--)
      dp[j] = max(dp[j], dp[j - pp->second] + pp->first);
  if (l == r) {
    long long ans = 0;
    for (int i = k; i; i--) ans = (ans * base + dp[i]) % MOD;
    printf("%lld\n", ans);
    return;
  }
  int mid = (l + r) >> 1;
  long long f[maxk];
  memcpy(f, dp, sizeof(f));
  solve(p << 1, l, mid, f);
  memcpy(f, dp, sizeof(f));
  solve(p << 1 | 1, mid + 1, r, f);
}
int main() {
  int n, q, opt, u, v, t = 1;
  n = qr();
  k = qr();
  for (int i = 1; i <= n; i++) {
    u = qr();
    v = qr();
    exb[i] = (exhibits){u, v, t, -1};
  }
  scanf("%d", &q);
  while (q--) {
    opt = qr();
    if (opt == 1) {
      u = qr();
      v = qr();
      exb[++n] = (exhibits){u, v, t, -1};
    } else if (opt == 2)
      exb[qr()].r = t - 1;
    else
      t++;
  }
  t--;
  for (int i = 1; i <= n; i++) {
    exb[i].r = exb[i].r == -1 ? t : exb[i].r;
    if (exb[i].l <= exb[i].r)
      add(1, 1, t, exb[i].l, exb[i].r, exb[i].v, exb[i].w);
  }
  solve(1, 1, t, DP);
  return 0;
}
inline int qr() {
  int num = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) num = (num << 1) + (num << 3) + ch - 48, ch = getchar();
  return num;
}
