Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, q, l, r, id, i, x, y;
long long nw;
long long p = 1e9 + 7;
long long a[200005];
long long w[200005];
long long t1[200005];
long long t2[200005];
int lowbit(int x) { return x & -x; }
void ud_t1(int x, long long dt) {
  int i;
  for (i = x; i <= n; i += lowbit(i)) t1[i] += dt;
}
void ud_t2(int x, long long dt) {
  int i;
  for (i = x; i <= n; i += lowbit(i)) t2[i] += dt, t2[i] = (t2[i] % p + p) % p;
}
long long query_t1(int x) {
  int i;
  long long ans = 1ll * 0;
  for (i = x; i; i -= lowbit(i)) ans += t1[i];
  return ans;
}
long long query_t2(int x) {
  int i;
  long long ans = 1ll * 0;
  for (i = x; i; i -= lowbit(i)) ans += t2[i], ans = (ans % p + p) % p;
  return ans;
}
long long sum_t1(int l, int r) { return query_t1(r) - query_t1(l - 1); }
long long sum_t2(int l, int r) {
  return ((query_t2(r) - query_t2(l - 1)) % p + p) % p;
}
long long find(int L, int R) {
  int l = L, r = R;
  int m;
  while (l < r) {
    m = (l + r) >> 1;
    if (sum_t1(L, m) >= sum_t1(m + 1, R))
      r = m;
    else
      l = m + 1;
  }
  return l;
}
int main() {
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 1; i <= n; i++) scanf("%lld", &w[i]);
  for (i = 1; i <= n; i++)
    ud_t1(i, w[i]), ud_t2(i, 1ll * w[i] * (a[i] - i) % p);
  long long loc;
  long long ans;
  while (q--) {
    scanf("%d%d", &x, &y);
    if (x < 0) {
      id = -x, nw = 1ll * y;
      long long dt1 = nw - w[id];
      w[id] = nw;
      ud_t1(id, dt1);
      long long dt2 = 1ll * dt1 * (a[id] - id) % p;
      ud_t2(id, dt2);
    } else {
      l = x, r = y;
      loc = find(l, r);
      ans = 1ll * (a[loc] - loc) % p * (sum_t1(l, loc - 1) % p) % p;
      ans -= 1ll * ((sum_t2(l, loc - 1) % p + p) % p);
      ans %= p;
      ans += 1ll * ((sum_t2(loc + 1, r) % p + p) % p);
      ans %= p;
      ans -= 1ll * (a[loc] - loc) % p * (sum_t1(loc + 1, r) % p) % p;
      ans = (ans % p + p) % p;
      printf("%lld\n", ans);
    }
  }
  return 0;
}