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;
namespace whatever {
int readu() {
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  int value = ch - '0';
  ch = getchar();
  while (isdigit(ch)) {
    value = value * 10 + ch - '0';
    ch = getchar();
  }
  return value;
}
int reads() {
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  int sign = 1;
  if (ch == '-') {
    sign = -1;
    ch = getchar();
  }
  int value = ch - '0';
  ch = getchar();
  while (isdigit(ch)) {
    value = value * 10 + ch - '0';
    ch = getchar();
  }
  return value * sign;
}
void writeu(int n) {
  if (n < 10)
    putchar(n + '0');
  else {
    writeu(n / 10);
    putchar(n % 10 + '0');
  }
}
const int mod = 1000000007;
int n;
void add(int &a, int b) {
  a += b;
  a %= mod;
}
void add(long long &a, long long b) { a += b; }
int low_bit(int a) { return a & -a; }
template <typename type>
void add(type *bit, int index, long long value) {
  while (index <= n) {
    add(bit[index], value);
    index += (index & -index);
  }
}
template <typename type>
type sum(type *bit, int index) {
  type result = 0;
  while (index != 0) {
    add(result, bit[index]);
    index -= (index & -index);
  }
  return result;
}
pair<int, long long> lower_bound(long long *bit, long long value) {
  int index = (1 << 18);
  for (int i = 17; i >= 0; --i)
    if (index - (1 << i) <= n && bit[index - (1 << i)] < value)
      value -= bit[index - (1 << i)];
    else
      index -= (1 << i);
  return make_pair(index, value);
}
void run() {
  n = readu();
  int q = readu();
  static int a[200000 + 1];
  static int w[200000 + 1];
  static long long bit_weight[200000 + 1];
  static int bit_prod[200000 + 1];
  for (int i = 1; i <= n; ++i) a[i] = readu();
  for (int i = 1; i <= n; ++i) {
    w[i] = readu();
    add(bit_weight, i, w[i]);
    add(bit_prod, i, (long long)(a[i] - i) * w[i] % mod);
  }
  while (--q != -1) {
    int x = reads();
    int y = readu();
    if (x < 0) {
      int index = -x;
      int new_w = y;
      add(bit_weight, index, new_w - w[index]);
      add(bit_prod, index,
          (long long)(a[index] - index) * (new_w - w[index] + mod) % mod);
      w[index] = new_w;
    } else {
      int l = x;
      int r = y;
      long long sum_before_l = sum(bit_weight, l - 1);
      long long sum_until_r = sum(bit_weight, r);
      long long sum_between = sum_until_r - sum_before_l;
      long long sum_half_way = sum_before_l + (sum_between + 1) / 2;
      auto [m, remain] = lower_bound(bit_weight, sum_half_way);
      assert(sum(bit_weight, m - 1) + remain == sum_half_way);
      int coeff = (a[m] - m + mod) % mod;
      long long sum_until_m = sum(bit_weight, m);
      int result =
          (sum_until_m * 2 - sum_before_l - sum_until_r) % mod * coeff % mod;
      sum_before_l = sum(bit_prod, l - 1);
      sum_until_m = sum(bit_prod, m);
      sum_until_r = sum(bit_prod, r);
      result = (result - (sum_until_m - sum_before_l) +
                (sum_until_r - sum_until_m)) %
               mod;
      result %= mod;
      result += mod;
      result %= mod;
      writeu(result);
      putchar('\n');
    }
  }
}
}  // namespace whatever
int main() { whatever::run(); }