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]. 
#include <bits/stdc++.h>
const int Z = 1 << 18;
const long long mod = 1000000007;
struct node {
  node() { a = w = s = 0; }
  node(long long a_, long long w_) {
    a = a_;
    w = w_;
    s = a * w % mod;
  }
  node(long long a_, long long w_, long long s_) {
    a = a_;
    w = w_;
    s = s_;
  }
  long long a, w, s;
  node operator+(const node &t) const {
    return {a + t.a, w + t.w, (s + t.s) % mod};
  }
} IT[Z * 2];
int N, Q, A[200200], W[200200];
void push(int x) {
  IT[x + Z] = node(A[x], W[x]);
  x = (x + Z) / 2;
  while (x) {
    IT[x] = IT[x * 2] + IT[x * 2 + 1];
    x /= 2;
  }
}
node get(int x, int y) {
  node r;
  x += Z;
  y += Z;
  while (x < y) {
    if (x & 1) r = r + IT[x++];
    if (y & 1) r = r + IT[--y];
    x /= 2;
    y /= 2;
  }
  return r;
}
int getPos(long long w) {
  int x = 1;
  while (x < Z) {
    x *= 2;
    if (w > IT[x].w) {
      w -= IT[x].w;
      x++;
    }
  }
  return x - Z;
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]), A[i] -= i;
  for (int i = 1; i <= N; i++) scanf("%d", &W[i]);
  for (int i = 1; i <= N; i++) IT[i + Z] = node(A[i], W[i]);
  for (int i = Z - 1; i >= 1; i--) IT[i] = IT[i * 2] + IT[i * 2 + 1];
  while (Q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      W[-x] = y;
      push(-x);
    } else {
      y++;
      node s = get(x, y);
      node l = get(0, x);
      long long want = l.w + (s.w + 1) / 2;
      int p = getPos(want);
      long long c = A[p];
      node a = get(x, p);
      node b = get(p + 1, y);
      long long as = (a.w % mod * c) + mod - a.s;
      long long bs = b.s + mod - (b.w % mod * c) % mod;
      printf("%lld\n", (as + bs) % mod);
    }
  }
  return 0;
}
