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 = 1e9 + 7;
const long long P = 1e7 + 19;
int n, q;
long long k;
struct obj {
  long long v;
  int w;
  int fir, en;
} e[40010];
int num, sf[30010];
vector<int> tree[30010 << 2];
void opt1(int o, int l, int r, int L, int R, int x) {
  if (l >= L && r <= R) {
    tree[o].push_back(x);
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= L) opt1(o << 1, l, mid, L, R, x);
  if (mid < R) opt1(o << 1 | 1, mid + 1, r, L, R, x);
}
long long dp[1010];
long long ksm(long long x, long long y) {
  long long maretu = 1;
  for (; y; y >>= 1) {
    if (y & 1) {
      maretu = maretu * x % MOD;
    }
    x = x * x % MOD;
  }
  return maretu;
}
long long mi[400010];
void dfs(int o, int l, int r) {
  long long now[1010];
  for (int i = 0; i < tree[o].size(); ++i) {
    for (int j = k; j >= e[tree[o][i]].w; --j) {
      now[j] = dp[j] = max(dp[j], dp[j - e[tree[o][i]].w] + e[tree[o][i]].v);
    }
  }
  memcpy(now, dp, sizeof(dp));
  if (l == r) {
    if (sf[l] == 0) return;
    long long ans = 0;
    for (int i = 1; i <= k; ++i) {
      ans += dp[i] * mi[i - 1];
      ans %= MOD;
    }
    printf("%lld\n", ans);
    return;
  }
  int mid = (l + r) >> 1;
  dfs(o << 1, l, mid);
  memcpy(dp, now, sizeof(now));
  dfs(o << 1 | 1, mid + 1, r);
}
int main() {
  scanf("%d%lld", &n, &k);
  mi[0] = 1;
  for (int i = 1; i <= k; ++i) {
    mi[i] = mi[i - 1] * P % MOD;
  }
  for (int i = 1; i <= n; ++i) {
    ++num;
    scanf("%lld%d", &e[num].v, &e[num].w);
    e[num].fir = 1;
  }
  scanf("%d", &q);
  for (int i = 1; i <= n + q; ++i) {
    e[i].en = q;
  }
  for (int i = 1; i <= q; ++i) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      ++num;
      scanf("%lld%d", &e[num].v, &e[num].w);
      e[num].fir = i;
    }
    if (opt == 2) {
      int x;
      scanf("%d", &x);
      e[x].en = i;
    }
    if (opt == 3) sf[i] = 1;
  }
  for (int i = 1; i <= num; ++i) {
    opt1(1, 1, q, e[i].fir, e[i].en, i);
  }
  dfs(1, 1, q);
}
