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]. 
#include <bits/stdc++.h>
using namespace std;
int n, X[201000], w[201000], Q;
long long BIT[201000], SS[201000], BIT2[201000], Mod = 1000000007, BIT3[201000];
void Add(int a, long long x) {
  while (a <= n) {
    BIT[a] += x;
    a += (a & -a);
  }
}
long long Sum(int a) {
  long long r = 0;
  while (a) {
    r += BIT[a];
    a -= (a & -a);
  }
  return r;
}
void Add2(int a, long long x) {
  while (a <= n) {
    BIT2[a] = (BIT2[a] + x) % Mod;
    a += (a & -a);
  }
}
long long Sum2(int a) {
  long long r = 0;
  while (a) {
    r += BIT2[a];
    a -= (a & -a);
  }
  return r % Mod;
}
void Add3(int a, long long x) {
  while (a <= n) {
    BIT3[a] = (BIT3[a] + x) % Mod;
    a += (a & -a);
  }
}
long long Sum3(int a) {
  long long r = 0;
  while (a) {
    r += BIT3[a];
    a -= (a & -a);
  }
  return r % Mod;
}
void Go(int b, int e) {
  long long G = Sum(e) - Sum(b - 1);
  long long BS = Sum(b - 1);
  int bb = b - 1, ee = e, mid, r = 0;
  while (bb <= ee) {
    mid = (bb + ee) >> 1;
    if ((Sum(mid) - BS) * 2 <= G)
      r = mid, bb = mid + 1;
    else
      ee = mid - 1;
  }
  r++;
  long long bs = Sum(r) - Sum(b - 1);
  long long es = Sum(e) - Sum(r);
  long long res = bs % Mod * X[r] - (Sum2(r) - Sum2(b - 1));
  res += (Sum2(e) - Sum2(r)) - es % Mod * X[r] % Mod;
  res -= bs % Mod * r % Mod - (Sum3(r) - Sum3(b - 1));
  res -= (Sum3(e) - Sum3(r)) - es % Mod * r % Mod;
  res = (res + Mod * 100) % Mod;
  printf("%lld\n", res);
}
int main() {
  int i;
  scanf("%d%d", &n, &Q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &X[i]);
    SS[i] = SS[i - 1] + X[i];
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    Add(i, w[i]);
    Add2(i, 1ll * w[i] * X[i] % Mod);
    Add3(i, 1ll * w[i] * i % Mod);
  }
  while (Q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a < 0) {
      a = -a;
      Add(a, b - w[a]);
      Add2(a, 1ll * (b - w[a] + Mod) * X[a] % Mod);
      Add3(a, 1ll * (b - w[a] + Mod) * a % Mod);
      w[a] = b;
      continue;
    }
    Go(a, b);
  }
}
