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;
long long mod = 1e9 + 7;
int norm(int a) {
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
int norm(const long long &a) { return norm(int(a % mod)); }
long long mul(long long a, long long n) { return ((a * n) % mod); }
void inc(vector<long long> &f, int pos, long long val) {
  int N = (int)f.size();
  for (; pos < N; pos |= pos + 1) f[pos] += val;
}
long long sum(vector<long long> &f, int pos) {
  long long ans = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ans += f[pos];
  return ans;
}
long long getSum(vector<long long> &f, int l, int r) {
  return sum(f, r - 1) - sum(f, l - 1);
}
long long getNSum(vector<long long> &f, int l, int r) {
  return norm(sum(f, r - 1) - sum(f, l - 1));
}
int main() {
  int n, q;
  cin >> n >> q;
  vector<long long> a(n), w(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> w[i];
  vector<long long> fn(n + 1), fnW(n + 1);
  for (int i = 0; i < n; i++) {
    a[i] -= i;
    inc(fn, i, w[i]);
    inc(fnW, i, mul(a[i], w[i]));
  }
  for (int t = 0; t < q; t++) {
    int first, second;
    cin >> first >> second;
    if (first < 0) {
      first = -first - 1;
      inc(fn, first, -w[first]);
      inc(fnW, first, -mul(a[first], w[first]));
      w[first] = second;
      inc(fn, first, w[first]);
      inc(fnW, first, mul(a[first], w[first]));
      continue;
    }
    first--;
    long long sum = getSum(fn, first, second);
    int left = first, right = second;
    while (right - left > 1) {
      int mid = left + (right - left) / 2;
      if (2 * getSum(fn, first, mid) > sum)
        right = mid;
      else
        left = mid;
    }
    int ans = norm(mul(a[left], getNSum(fn, first, left)) -
                   getNSum(fnW, first, left));
    ans = norm(ans + norm(getNSum(fnW, left, second) -
                          mul(a[left], getNSum(fn, left, second))));
    cout << ans << endl;
  }
  return 0;
}