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 long long INF = 1e18 + 7;
const long long ABS = 1e9 + 7;
struct Fenv {
  vector<long long> T;
  void init(int n) { T.resize(n); }
  void add(int i, long long v, bool f) {
    for (; i < T.size(); i = (i | (i + 1)))
      if (f)
        T[i] = (T[i] + v) % ABS;
      else
        T[i] = T[i] + v;
  }
  long long sum(int r) {
    long long res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) res = res + T[r];
    return res;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
  long long fi(int i) { return sum(i, i); }
};
long long posn(long long x) {
  if (x >= ABS) x -= ABS;
  if (x < 0) x += ABS;
  return x;
}
int main() {
  int n, q;
  cin >> n >> q;
  Fenv Fwa, S;
  vector<long long> a(n), w(n);
  Fwa.init(n), S.init(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    S.add(i, w[i], 0);
    Fwa.add(i, w[i] * a[i], 1);
  }
  for (; q > 0; --q) {
    int l, r;
    cin >> l >> r;
    if (l < 0) {
      l *= -1;
      l--;
      long long cw = S.fi(l);
      S.add(l, r - cw, 0);
      cw = Fwa.fi(l);
      Fwa.add(l, r * a[l] - cw, 1);
    } else {
      l--, r--;
      if (l == r) {
        cout << 0 << endl;
        continue;
      }
      long long SM = S.sum(l, r);
      int L = l, R = r + 1;
      while (L + 1 < R) {
        int m = (L + R) / 2;
        if (2 * S.sum(l, m) > SM)
          R = m;
        else
          L = m;
      }
      int k;
      if (2 * S.sum(l, L) < SM)
        k = R;
      else
        k = L;
      long long ans = 0;
      ans = posn(a[k] * S.sum(l, k) - Fwa.sum(l, k));
      ans = posn(ans + Fwa.sum(k, r) - a[k] * S.sum(k, r));
      ans = posn(ans % ABS);
      cout << ans << endl;
    }
  }
  return 0;
}