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;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char ch[4000010];
inline string get() {
  scanf("%s", ch);
  return string(ch);
}
inline void read(long long *a, long long n) {
  for (long long i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
template <class blank, class blank2>
inline void smin(blank &a, blank2 b) {
  a = a <= b ? a : b;
}
template <class blank, class blank2>
inline void smax(blank &a, blank2 b) {
  a = a >= b ? a : b;
}
const long long maxn = 2e5 + 10;
const long long maxm = 4e6 + 10;
const long long maxlg = 22;
const long long base = 29;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
const double eps = 1e-4;
const long long SQ = 317;
struct st {
  long long l, r, v, w;
} tblo[maxn];
long long n, k;
long long dpcpy[maxn], dp[maxn], asked, res[maxn];
vector<pair<long long, long long> > stree[4 * maxn];
const long long p = 1e7 + 19;
inline void update(st &shit, long long p = 0, long long l = 0,
                   long long r = asked - 1) {
  if (l > shit.r | r < shit.l || shit.l > shit.r) return;
  if (l >= shit.l && r <= shit.r) {
    stree[p].push_back({shit.v, shit.w});
    return;
  }
  long long mid = (l + r) >> 1;
  update(shit, 2 * p + 1, l, mid), update(shit, 2 * p + 2, mid + 1, r);
}
inline void calc_dp(long long v, long long w) {
  for (long long i = k; i >= w; i--) dpcpy[i] = max(dpcpy[i], dpcpy[i - w] + v);
}
inline void build(long long p = 0, long long l = 0, long long r = asked - 1) {
  long long pre[maxn];
  for (long long i = 0; i <= k; i++) pre[i] = dpcpy[i];
  for (auto u : stree[p]) calc_dp(u.first, u.second);
  if (l == r) {
    long long tavan = 1;
    long long val = dpcpy[1];
    for (long long m = 1; m <= k;
         m++, tavan = tavan * ::p % mod, smax(val, dpcpy[m])) {
      if (val > 0) res[l] = (res[l] + val * tavan) % mod;
    }
    for (long long i = 0; i <= k; i++) dpcpy[i] = pre[i];
    return;
  }
  long long mid = (l + r) >> 1;
  build(2 * p + 1, l, mid);
  build(2 * p + 2, mid + 1, r);
  for (long long i = 0; i <= k; i++) dpcpy[i] = pre[i];
}
int32_t main() {
  n = in(), k = in();
  memset(dpcpy, -63, sizeof dpcpy);
  dpcpy[0] = 0;
  for (long long i = 0; i < n; i++)
    tblo[i].l = 0, tblo[i].r = inf, tblo[i].v = in(), tblo[i].w = in();
  long long q = in();
  while (q--) {
    long long tp = in();
    if (tp == 1) {
      tblo[n].v = in(), tblo[n].w = in(), tblo[n].l = asked, tblo[n].r = inf;
      n++;
    } else if (tp == 2) {
      long long id = in() - 1;
      tblo[id].r = asked - 1;
    } else
      asked++;
  }
  for (long long i = 0; i < n; i++) {
    smin(tblo[i].r, asked - 1);
    update(tblo[i]);
  }
  build();
  for (long long i = 0; i < asked; i++, cout << "\n") cout << res[i];
}
