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 IO {
const int BUFFER_SIZE = 4096;
char input_buffer[BUFFER_SIZE];
int input_pos = BUFFER_SIZE;
int input_read = BUFFER_SIZE;
char output_buffer[BUFFER_SIZE];
int output_pos = 0;
uint8_t lookup[100];
void _update_input_buffer() {
  if (input_pos >= input_read && input_read == BUFFER_SIZE) {
    input_read = fread(input_buffer, sizeof(char), BUFFER_SIZE, stdin);
    input_pos = 0;
  }
}
char next_char(bool advance = true) {
  _update_input_buffer();
  return input_buffer[advance ? input_pos++ : input_pos];
}
template <typename T>
void read_int(T &number) {
  while (isspace(next_char(false))) next_char();
  bool negative = next_char(false) == '-';
  if (negative) next_char();
  number = 0;
  while (isdigit(next_char(false))) number = 10 * number + (next_char() - '0');
  if (negative) number = -number;
}
void flush_output() {
  fwrite(output_buffer, sizeof(char), output_pos, stdout);
  output_pos = 0;
}
void write_char(char c) {
  if (output_pos == BUFFER_SIZE) flush_output();
  output_buffer[output_pos++] = c;
}
template <typename T>
void write_int(T number, char after = 0) {
  if (number < 0) {
    write_char('-');
    number = -number;
  }
  static char number_buffer[100];
  int length = 0;
  do {
    if (number < 10) {
      number_buffer[length++] = number + '0';
      number = 0;
    } else {
      uint8_t lookup_value = lookup[number % 100];
      number /= 100;
      number_buffer[length++] = (lookup_value & 15) + '0';
      number_buffer[length++] = (lookup_value >> 4) + '0';
    }
  } while (number > 0);
  for (int i = length - 1; i >= 0; i--) write_char(number_buffer[i]);
  if (after) write_char(after);
}
void init() {
  bool exit_success = atexit(flush_output) == 0;
  assert(exit_success);
  for (int i = 0; i < 100; i++) lookup[i] = (i / 10 << 4) + i % 10;
}
}  // namespace IO
const int MOD = 1e9 + 7;
template <typename T1, typename T2>
void mod_add(T1 &a, T2 b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
struct tree_node {
  long long w_sum;
  int aw_sum;
  tree_node() : w_sum(0), aw_sum(0) {}
  void join(const tree_node &other) {
    w_sum += other.w_sum;
    mod_add(aw_sum, other.aw_sum);
  }
};
int N, LOG, Q;
vector<int> A, W;
vector<tree_node> tree;
void build_tree() {
  for (int i = 1; i <= N; i++) {
    for (int k = (i & -i) >> 1; k > 0; k >>= 1) tree[i].join(tree[i - k]);
    tree[i].w_sum += W[i - 1];
    tree[i].aw_sum = (tree[i].aw_sum + (long long)W[i - 1] * A[i - 1]) % MOD;
  }
}
void tree_update(int index, long long w_change, int a) {
  int aw_change = (MOD + w_change) * a % MOD;
  for (int i = index + 1; i <= N; i += i & -i) {
    tree[i].w_sum += w_change;
    mod_add(tree[i].aw_sum, aw_change);
  }
}
tree_node tree_query(int count) {
  tree_node answer;
  for (int i = count; i > 0; i -= i & -i) answer.join(tree[i]);
  return answer;
}
int solve(int start, int end) {
  tree_node tree_end = tree_query(end);
  tree_node tree_start = tree_query(start);
  long long search = (tree_end.w_sum + tree_start.w_sum + 1) / 2;
  int goal = 0;
  long long w_sum = 0, aw_sum = 0;
  for (int k = LOG - 1; k >= 0; k--)
    if (goal + (1 << k) <= N && search > w_sum + tree[goal + (1 << k)].w_sum) {
      goal += 1 << k;
      w_sum += tree[goal].w_sum;
      aw_sum += tree[goal].aw_sum;
    }
  assert(start <= goal && goal < end);
  long long result =
      ((w_sum - tree_start.w_sum) - (tree_end.w_sum - w_sum)) % MOD * A[goal] +
      (tree_end.aw_sum - aw_sum) - (aw_sum - tree_start.aw_sum);
  result %= MOD;
  mod_add(result, MOD);
  return result;
}
int main() {
  IO::init();
  IO::read_int(N);
  IO::read_int(Q);
  A.resize(N);
  W.resize(N);
  LOG = 32 - __builtin_clz(N);
  assert(1 << LOG > N);
  tree.resize(N + 1);
  for (int i = 0; i < N; i++) {
    IO::read_int(A[i]);
    A[i] -= i;
  }
  for (int i = 0; i < N; i++) IO::read_int(W[i]);
  build_tree();
  for (int q = 0; q < Q; q++) {
    int x, y;
    IO::read_int(x);
    IO::read_int(y);
    if (x < 0) {
      x = -x - 1;
      tree_update(x, y - W[x], A[x]);
      W[x] = y;
    } else {
      IO::write_int(solve(x - 1, y), '\n');
    }
  }
}