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;
long long n, q, A[200010], W[200010], Sm1[800010], Sm2[800010];
void upd(long long x) {
  Sm1[x] = (Sm1[x * 2] + Sm2[x * 2 + 1]) % 1000000007;
  Sm2[x] = (Sm2[x * 2] + Sm2[x * 2 + 1]) % 1000000007;
}
void add1(long long x, long long v) {
  for (long long i = x; i <= n; i += i & -i) Sm1[i] = Sm1[i] + v;
}
void add2(long long x, long long v) {
  for (long long i = x; i <= n; i += i & -i) Sm2[i] = (Sm2[i] + v) % 1000000007;
}
long long ask1(long long x) {
  long long ans = 0;
  for (long long i = x; i; i -= i & -i) ans = ans + Sm1[i];
  return ans;
}
long long ask2(long long x) {
  long long ans = 0;
  for (long long i = x; i; i -= i & -i) ans = (ans + Sm2[i]) % 1000000007;
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &A[i]);
  for (long long i = 1; i <= n; i++) scanf("%lld", &W[i]);
  for (long long i = 1; i <= n; i++) {
    add1(i, W[i]);
    add2(i, (A[i] - i) * W[i] % 1000000007);
  }
  for (long long T = 1; T <= q; T++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if (a > 0) {
      long long o = ask1(a - 1), p = ask1(b), v = p - o;
      long long ans = a;
      for (long long l = a, r = b; l <= r;) {
        long long md = (l + r) / 2;
        if ((ask1(md) - o) * 2 >= v)
          ans = md, r = md - 1;
        else
          l = md + 1;
      }
      long long tot =
          (A[ans] - ans) * ((ask1(ans) - ask1(a - 1)) % 1000000007) -
          (ask2(ans) - ask2(a - 1));
      tot = tot + (ask2(b) - ask2(ans)) -
            (A[ans] - ans) * ((ask1(b) - ask1(ans)) % 1000000007);
      printf("%lld\n", (tot % 1000000007 + 1000000007) % 1000000007);
    } else {
      a = -a;
      add1(a, -W[a]);
      add2(a, -(A[a] - a) * W[a] % 1000000007);
      W[a] = b;
      add1(a, W[a]);
      add2(a, (A[a] - a) * W[a] % 1000000007);
    }
  }
  return 0;
}