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 maxn = 1e5 + 500;
const long long maxq = 3e4 + 500;
const long long maxw = 1004;
const long long mod = 1e9 + 7;
const long long inf = 1e16;
const long long base = 1e7 + 19;
long long tav[maxn];
long long v[maxn];
long long w[maxn];
long long st[maxn];
long long et[maxn];
long long typ[maxn];
long long cnt;
vector<long long> vec[maxn * 4];
bool soal[maxn * 4];
void add(long long L, long long R, long long node, long long l, long long r,
         long long x) {
  if (l <= L && R <= r) {
    vec[node].push_back(x);
    return;
  }
  if (R <= l || r <= L) {
    return;
  }
  long long mid = (L + R) / 2;
  add(L, mid, 2 * node, l, r, x);
  add(mid, R, 2 * node + 1, l, r, x);
}
void ad(long long l, long long r, long long node, long long x) {
  if (l + 1 == r) {
    soal[node] = 1;
    return;
  }
  long long mid = (l + r) / 2;
  if (x < mid) {
    ad(l, mid, 2 * node, x);
  } else {
    ad(mid, r, 2 * node + 1, x);
  }
}
struct napsak {
  long long dp[maxw];
};
long long k;
stack<napsak> stk;
void dfs(long long l, long long r, long long node) {
  for (auto i : vec[node]) {
    long long vv = v[i];
    long long ww = w[i];
    for (long long j = k; j >= ww; j--) {
      stk.top().dp[j] = max(stk.top().dp[j], stk.top().dp[j - ww] + vv);
    }
  }
  if (l + 1 == r) {
    if (soal[node]) {
      long long ans = 0;
      for (long long i = 1; i <= k; i++) {
        ans = (ans + stk.top().dp[i] * tav[i - 1]) % mod;
      }
      cout << ans << endl;
    }
    stk.pop();
    return;
  }
  long long mid = (l + r) / 2;
  stk.push(stk.top());
  dfs(l, mid, 2 * node);
  stk.push(stk.top());
  dfs(mid, r, 2 * node + 1);
  stk.pop();
}
int main() {
  tav[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    tav[i] = (tav[i - 1] * base) % mod;
  }
  long long n;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i] >> w[i];
  }
  long long q;
  cin >> q;
  for (long long i = 1; i <= n; i++) {
    st[i] = 0;
    et[i] = q;
  }
  cnt = n + 1;
  for (long long i = 0; i < q; i++) {
    cin >> typ[i];
    if (typ[i] == 1) {
      long long vv, ww;
      cin >> vv >> ww;
      v[cnt] = vv;
      w[cnt] = ww;
      st[cnt] = i;
      et[cnt] = q;
      cnt++;
    }
    if (typ[i] == 2) {
      long long x;
      cin >> x;
      et[x] = i;
    }
    if (typ[i] == 3) {
      ad(0, q, 1, i);
    }
  }
  napsak a;
  for (long long i = 0; i <= k; i++) {
    a.dp[i] = 0;
  }
  for (long long i = 0; i < cnt; i++) {
    add(0, q, 1, st[i], et[i], i);
  }
  stk.push(a);
  dfs(0, q, 1);
}
