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;
const int N = 400010;
int n, Q;
long long a[N], w[N], Tree[N * 4], sum[N * 4];
void up(int x) {
  Tree[x] = Tree[x << 1] + Tree[x << 1 | 1];
  sum[x] = (sum[x << 1] + sum[x << 1 | 1]) % 1000000007;
}
void build(int l, int r, int id) {
  if (l == r) {
    Tree[id] = w[l];
    sum[id] = (long long)(a[l] - l) * w[l] % 1000000007;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, id << 1);
  build(mid + 1, r, id << 1 | 1);
  up(id);
}
void change(int l, int r, int id, int x, int y) {
  if (l == r) {
    Tree[id] = y;
    sum[id] = (long long)(a[l] - l) * y % 1000000007;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    change(l, mid, id << 1, x, y);
  else
    change(mid + 1, r, id << 1 | 1, x, y);
  up(id);
}
long long query_sum(int l, int r, int id, int x, int y) {
  if (l > y || r < x || Tree[id] == 0) return 0;
  if (x <= l && r <= y) return Tree[id];
  int mid = (l + r) >> 1;
  return query_sum(l, mid, id << 1, x, y) +
         query_sum(mid + 1, r, id << 1 | 1, x, y);
}
long long query(int l, int r, int id, int x, int y) {
  if (l > y || r < x) return 0;
  if (x <= l && r <= y) return sum[id];
  int mid = (l + r) >> 1;
  return (query(l, mid, id << 1, x, y) + query(mid + 1, r, id << 1 | 1, x, y)) %
         1000000007;
}
int erfen(int l, int r) {
  int L = l;
  long long sum = query_sum(1, n, 1, l, r);
  while (l < r) {
    int mid = (l + r) >> 1;
    if (query_sum(1, n, 1, L, mid) * 2 >= sum)
      r = mid;
    else
      l = mid + 1;
  }
  return r;
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%I64d", &w[i]);
  build(1, n, 1);
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      change(1, n, 1, x, y);
    } else {
      int p = erfen(x, y);
      long long cnt1 = query_sum(1, n, 1, x, p) % 1000000007,
                cnt2 = query_sum(1, n, 1, p, y) % 1000000007;
      long long s1 =
          ((long long)(a[p] - p) * cnt1 % 1000000007 - query(1, n, 1, x, p)) %
          1000000007;
      long long s2 =
          (query(1, n, 1, p, y) - (long long)(a[p] - p) * cnt2 % 1000000007) %
          1000000007;
      printf("%I64d\n", ((s1 + s2) % 1000000007 + 1000000007) % 1000000007);
    }
  }
  return 0;
}