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 P = 1e7 + 19;
const int MOD = 1e9 + 7;
const int maxn = 44444;
const int maxh = 18;
int m = 0, n, K, x, Q, type, v[maxn], w[maxn], l[maxn], r[maxn], ans[maxn];
long long dp[maxn], backup[maxh][maxn];
vector<int> tr[maxn * 2];
inline int getnode(int l, int r) { return (l + r) | (l != r); }
void add(int l, int r, int a, int b, int id) {
  if (a > r || b < l) return;
  int k = getnode(l, r);
  if (a <= l && r <= b) {
    tr[k].push_back(id);
    return;
  }
  int mid = l + r >> 1;
  add(l, mid, a, b, id);
  add(mid + 1, r, a, b, id);
}
void dfs(int l, int r, int dep) {
  int k = getnode(l, r);
  for (int i = 0; i <= K; ++i) backup[dep][i] = dp[i];
  for (int i = 0; i < tr[k].size(); ++i) {
    int x = tr[k][i];
    for (int j = K; j >= w[x]; --j) dp[j] = max(dp[j], dp[j - w[x]] + v[x]);
  }
  if (l == r) {
    ans[l] = 0;
    int tmp = 1;
    for (int i = 1; i <= K; ++i, tmp = (long long)tmp * P % MOD)
      ans[l] = ((long long)dp[i] % MOD * tmp + ans[l]) % MOD;
    for (int i = 0; i <= K; ++i) dp[i] = backup[dep][i];
    return;
  }
  int mid = l + r >> 1;
  dfs(l, mid, dep + 1);
  dfs(mid + 1, r, dep + 1);
  for (int i = 0; i <= K; ++i) dp[i] = backup[dep][i];
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &v[i], &w[i]);
  scanf("%d", &Q);
  for (int i = 1; i <= n; ++i) l[i] = 1, r[i] = Q;
  for (int i = 1; i <= Q; ++i) {
    scanf("%d", &type);
    if (type == 1) {
      n++;
      scanf("%d%d", &v[n], &w[n]);
      l[n] = m + 1;
      r[n] = Q;
    }
    if (type == 2) {
      scanf("%d", &x);
      r[x] = m;
    }
    if (type == 3) {
      m++;
    }
  }
  for (int i = 1; i <= n; ++i) r[i] = min(r[i], m);
  for (int i = 1; i <= n; ++i) {
    if (l[i] <= r[i]) add(1, m, l[i], r[i], i);
  }
  memset(dp, 0, sizeof(dp));
  dfs(1, m, 1);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}
