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 modn = 1000000007;
const long long B = 10000019;
;
inline long long mod(long long x) { return x % modn; }
int n;
int K;
const int N = 112345;
int w[N], v[N], qu[N], ot[N], p[N];
int dp[15009][1009];
int cur;
inline void add(int v, int w) {
  for (int i = 0; i <= K; i++) {
    dp[cur][i] = (cur ? dp[cur - 1][i] : 0);
    if (i >= w)
      dp[cur][i] = max(dp[cur][i], (cur ? dp[cur - 1][i - w] : 0) + v);
  }
  cur++;
}
void solve(int l, int r) {
  if (l == r) {
    long long ans = 0;
    for (int i = K; i >= 1; i--)
      ans = mod(ans * B + (cur ? dp[cur - 1][i] : 0));
    printf("%lld\n", ans);
    return;
  }
  int m = (l + r) / 2;
  int ocur = cur;
  if (qu[m] - (l ? qu[l - 1] : 0)) {
    for (int i = m + 1; i <= r; i++)
      if (qu[i] - (i ? qu[i - 1] : 0) == 0 && ot[i] < l)
        add(v[ot[i]], w[ot[i]]);
    solve(l, m);
  }
  cur = ocur;
  if (qu[r] - qu[m]) {
    for (int i = l; i <= m; i++)
      if (qu[i] - (i ? qu[i - 1] : 0) == 0 && ot[i] > r) add(v[i], w[i]);
    solve(m + 1, r);
  }
  cur = ocur;
}
int main() {
  int i, q;
  scanf("%d %d", &n, &K);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &v[i], &w[i]);
    p[i + 1] = i;
    ot[i] = 1e8;
  }
  scanf("%d", &q);
  int ct = n;
  for (i = n; i < n + q; i++) {
    qu[i] = qu[i - 1];
    char t;
    scanf(" %c", &t);
    if (t == '1') {
      scanf("%d %d", &v[i], &w[i]);
      p[++ct] = i;
      ot[i] = 1e8;
    } else if (t == '2') {
      int x;
      scanf("%d", &x);
      ot[p[x]] = i;
      ot[i] = p[x];
    } else {
      qu[i]++;
    }
  }
  solve(0, n + q - 1);
}
