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 maxn = 1000020;
const int maxm = 1000020;
const int MOd = 1e9 + 7;
int a, ar[maxn], w[maxn], q;
long long fw[maxn];
int val[maxn][2];
int mul(int x, int y) { return (long long)x * y % MOd; }
int add(int a, int b) {
  a += b;
  a %= MOd;
  if (a < 0) a += MOd;
  return a;
}
void uval(int x, int y, int tp) {
  for (int i = x; i <= a; i += i & (-i)) val[i][tp] = add(val[i][tp], y);
}
long long fval(int l, int r, int tp) {
  long long ret = 0;
  for (int i = r; i > 0; i -= i & (-i)) ret = add(ret, val[i][tp]);
  for (int i = l - 1; i > 0; i -= i & (-i)) ret = add(ret, -val[i][tp]);
  return ret;
}
void up(int x, int y) {
  for (int i = x; i <= a; i += i & (-i)) fw[i] += y;
}
long long f(int l, int r) {
  long long ret = 0;
  for (int i = r; i > 0; i -= i & (-i)) ret += fw[i];
  for (int i = l - 1; i > 0; i -= i & (-i)) ret -= fw[i];
  return ret;
}
int main() {
  scanf("%d %d", &a, &q);
  for (int i = 1; i <= a; i++) {
    scanf("%d", &ar[i]);
  }
  for (int i = 1; i <= a; i++) {
    scanf("%d", &w[i]), up(i, w[i]);
    uval(i, mul(w[i], ar[i]), 0);
    uval(i, mul(w[i], i), 1);
  }
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x > 0) {
      int l = x, r = y, ans = 0;
      long long tot = f(x, y) / 2;
      while (l < r) {
        int m = (l + r) / 2;
        if (f(x, m) > tot)
          r = m;
        else
          l = m + 1;
      }
      ans = add(ans, add(mul(f(x, r) % MOd, ar[r]), -fval(x, r, 0)));
      ans = add(ans, -add(mul(f(x, r) % MOd, r), -fval(x, r, 1)));
      ans = add(ans, add(fval(r, y, 0), -mul(f(r, y) % MOd, ar[r])));
      ans = add(ans, -add(fval(r, y, 1), -mul(f(r, y) % MOd, r)));
      cout << ans << endl;
    } else {
      x = -x;
      up(x, y - w[x]);
      uval(x, mul(y - w[x], ar[x]), 0);
      uval(x, mul(y - w[x], x), 1);
      w[x] = y;
    }
  }
  return 0;
}