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 mxN = 2e5, M = 1e9 + 7;
int n, q, x, y;
long long a[mxN], w[mxN], ft[mxN + 1][3];
void upd(int i, long long x, int j) {
  for (++i; i <= n; i += i & -i) {
    ft[i][j] += x;
    if (j) ft[i][j] %= M;
  }
}
long long qry(int i, int j) {
  long long r = 0;
  for (; i; i -= i & -i) {
    r += ft[i][j];
    if (j) r %= M;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    cin >> w[i];
    upd(i, w[i], 0);
    upd(i, i * w[i], 1);
    upd(i, a[i] * w[i], 2);
  }
  while (q--) {
    cin >> x >> y;
    if (x < 0) {
      x = -x - 1;
      upd(x, y - w[x], 0);
      upd(x, x * (y - w[x]), 1);
      upd(x, a[x] * (y - w[x]), 2);
      w[x] = y;
    } else {
      --x;
      --y;
      long long s1 = qry(x, 0), s2 = qry(y + 1, 0);
      int lb = x, rb = y;
      while (lb <= rb) {
        int mb = (lb + rb) / 2;
        long long c = qry(mb + 1, 0);
        if (2 * (c - s1) >= s2 - s1)
          rb = mb - 1;
        else
          lb = mb + 1;
      }
      long long ans = 0;
      ans += (qry(y + 1, 2) - 2 * qry(lb, 2) + qry(x, 2)) -
             (qry(y + 1, 1) - 2 * qry(lb, 1) + qry(x, 1)) -
             (qry(y + 1, 0) - 2 * qry(lb, 0) + qry(x, 0)) % M * (a[lb] - lb);
      cout << (ans % M + M) % M << "\n";
    }
  }
}