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 int MAXN = 200100;
const int P = (1 << 18);
const int MAXP = 2 * P + 10;
const int MOD = 1e9 + 7;
namespace IO {
const int BUFFER_SIZE = 4096;
const int NUMBER_BUFFER = 100;
FILE *input_stream = stdin;
FILE *output_stream = stdout;
char input_buffer[BUFFER_SIZE];
int input_pos = BUFFER_SIZE;
int input_read = BUFFER_SIZE;
char output_buffer[BUFFER_SIZE];
int output_pos = 0;
char number_buffer[NUMBER_BUFFER];
void init(const char *input_file, const char *output_file) {
  input_stream = fopen(input_file, "r");
  output_stream = fopen(output_file, "w");
}
void _update_input_buffer() {
  if (input_pos >= input_read && input_read == BUFFER_SIZE) {
    input_read = fread(input_buffer, sizeof(char), BUFFER_SIZE, input_stream);
    input_pos = 0;
  }
}
char next_char(bool advance = true) {
  _update_input_buffer();
  return input_buffer[advance ? input_pos++ : input_pos];
}
bool on_whitespace() { return isspace(next_char(false)); }
bool on_digit() { return isdigit(next_char(false)); }
void skip_whitespace() {
  while (on_whitespace()) next_char();
}
bool input_finished() {
  skip_whitespace();
  return input_pos >= input_read && input_read < BUFFER_SIZE;
}
template <typename T>
void read_int(T &number) {
  skip_whitespace();
  bool negative = next_char(false) == '-';
  if (negative) next_char();
  number = 0;
  while (on_digit()) number = 10 * number + (next_char() - '0');
  if (negative) number = -number;
}
void flush_output() {
  fwrite(output_buffer, sizeof(char), output_pos, output_stream);
  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;
  }
  int length = 0;
  do {
    number_buffer[length++] = number % 10 + '0';
    number /= 10;
  } while (number > 0);
  for (int i = length - 1; i >= 0; i--) write_char(number_buffer[i]);
  if (after) write_char(after);
}
struct Dummy {
  ~Dummy() { flush_output(); }
} dummy;
}  // namespace IO
int N, Q;
long long a[MAXN];
long long w[MAXN];
long long wseg[MAXP], waseg[MAXP];
void upd(long long *seg, int cloc, long long cdiff) {
  cloc += P;
  while (cloc) {
    seg[cloc] += cdiff;
    cloc /= 2;
  }
}
long long ctot, dtot;
long long add(int cloc) {
  long long res = 0;
  cloc += P;
  for (int i = 0; i < 18; i++) {
    if (cloc & 1) {
      res += wseg[cloc - 1];
      dtot -= waseg[cloc - 1];
    }
    cloc /= 2;
  }
  return res;
}
int wsearch(long long sgoal) {
  int cloc = 1;
  ctot = 0;
  for (int i = 0; i < 18; i++) {
    cloc *= 2;
    if (wseg[cloc] + ctot < sgoal) {
      ctot += wseg[cloc];
      dtot += waseg[cloc] << 1;
      cloc++;
    }
  }
  return cloc - P;
}
int main() {
  IO::read_int(N);
  IO::read_int(Q);
  for (int i = 0; i < N; i++) {
    int x;
    IO::read_int(x);
    a[i] = x;
  }
  for (int i = 0; i < N; i++) {
    int x;
    IO::read_int(x);
    w[i] = x;
  }
  for (int i = 0; i < N; i++) {
    wseg[i + P] = w[i];
    waseg[i + P] = (a[i] - i) * w[i] % MOD;
  }
  for (int i = P - 1; i >= 0; i--) {
    wseg[i] = wseg[2 * i] + wseg[2 * i + 1];
    waseg[i] = waseg[2 * i] + waseg[2 * i + 1];
  }
  for (int i = 0; i < Q; i++) {
    int l, r;
    IO::read_int(l);
    IO::read_int(r);
    if (l < 0) {
      int cloc = -l - 1;
      long long cdiff = r - w[cloc];
      upd(wseg, cloc, cdiff);
      upd(waseg, cloc, ((a[cloc] - cloc) * cdiff) % MOD);
      w[cloc] = r;
    } else {
      l--;
      dtot = 0;
      long long sleft = add(l), sright = add(r);
      long long sgoal = (sleft + sright + 1) / 2;
      int mm = wsearch(sgoal);
      long long ws = 2 * ctot - sleft - sright;
      long long res = ws * (a[mm] - mm) - dtot;
      res = res % MOD + MOD;
      if (res >= MOD) res -= MOD;
      IO::write_int((int)res, '\n');
    }
  }
}