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, x[200005], at;
long long y[19][200005], sum, z[19][200005];
int ta, tb;
long long f1(int a, int b, int c, int d) {
  if (((1 << (c)) * (d)) == a && ((1 << (c)) * (d + 1)) == b) return y[c][d];
  c--;
  d = d * 2 + 1;
  if (b <= ((1 << (c)) * (d))) return f1(a, b, c, d - 1);
  if (a >= ((1 << (c)) * (d))) return f1(a, b, c, d);
  return f1(a, ((1 << (c)) * (d)), c, d - 1) + f1(((1 << (c)) * (d)), b, c, d);
}
void f2(int c, int d) {
  if (c == 0) {
    at = d;
    return;
  }
  c--;
  d = d * 2 + 1;
  if (sum < y[c][d - 1])
    f2(c, d - 1);
  else {
    sum -= y[c][d - 1];
    f2(c, d);
  }
}
long long f3(int a, int b, int c, int d) {
  if (((1 << (c)) * (d)) == a && ((1 << (c)) * (d + 1)) == b) return z[c][d];
  c--;
  d = d * 2 + 1;
  if (b <= ((1 << (c)) * (d))) return f3(a, b, c, d - 1);
  if (a >= ((1 << (c)) * (d))) return f3(a, b, c, d);
  return f3(a, ((1 << (c)) * (d)), c, d - 1) + f3(((1 << (c)) * (d)), b, c, d);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", x + i);
    x[i] -= i;
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", y[0] + i);
    z[0][i] = y[0][i] * x[i] % 1000000007;
  }
  for (int i = 1; i < 19; i++)
    for (int j = 0; ((1 << (i)) * (j)) < n; j++) {
      y[i][j] = y[i - 1][j * 2] + y[i - 1][j * 2 + 1];
      z[i][j] = z[i - 1][j * 2] + z[i - 1][j * 2 + 1];
    }
  while (q--) {
    scanf("%d%d", &ta, &tb);
    if (ta < 0) {
      ta = -ta - 1;
      y[0][ta] = tb;
      z[0][ta] = y[0][ta] * x[ta] % 1000000007;
      for (int i = 1; i < 19; i++) {
        ta >>= 1;
        y[i][ta] = y[i - 1][ta * 2] + y[i - 1][ta * 2 + 1];
        z[i][ta] = z[i - 1][ta * 2] + z[i - 1][ta * 2 + 1];
      }
    } else {
      ta--;
      sum = f1(ta, tb, 18, 0) / 2;
      if (ta) sum += f1(0, ta, 18, 0);
      f2(18, 0);
      sum = 0;
      if (ta < at)
        sum = (sum + f1(ta, at, 18, 0) % 1000000007 * x[at] % 1000000007 +
               1000000007 - f3(ta, at, 18, 0) % 1000000007) %
              1000000007;
      if (at < tb)
        sum = (sum + f3(at, tb, 18, 0) % 1000000007 + 1000000007 -
               f1(at, tb, 18, 0) % 1000000007 * x[at] % 1000000007) %
              1000000007;
      printf("%lld\n", sum);
    }
  }
  return 0;
}