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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int read() {
  char ch;
  bool flag = 0;
  int res;
  while (!isdigit(
      ch = (p1 == p2 &&
                    (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++)))
    (ch == '-') && (flag = true);
  for (res = ch - '0'; isdigit(
           ch = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin),
                             p1 == p2)
                     ? EOF
                     : *p1++));
       res = res * 10 + ch - '0')
    ;
  (flag) && (res = -res);
  return res;
}
char sr[1 << 21], z[20];
int C = -1, Z;
inline void Ot() { fwrite(sr, 1, C + 1, stdout), C = -1; }
inline void print(int x) {
  if (C > 1 << 20) Ot();
  if (x < 0) sr[++C] = 45, x = -x;
  while (z[++Z] = x % 10 + 48, x /= 10)
    ;
  while (sr[++C] = z[Z], --Z)
    ;
  sr[++C] = '\n';
}
const int N = 2e5 + 5;
const long long P = 1e9 + 7;
int n, q;
long long a[N], w[N];
namespace SUM {
long long c[N];
inline void add(int x, long long y) {
  for (; x <= n; x += x & -x) c[x] += y;
}
inline long long que(int x) {
  long long res = 0;
  for (; x; x -= x & -x) res += c[x];
  return res;
}
inline long long query(int l, int r) {
  if (r < l) return 0;
  return que(r) - que(l - 1);
}
}  // namespace SUM
namespace MUL {
long long c[N];
inline void add(int x, long long y) {
  y %= P;
  for (; x <= n; x += x & -x) (c[x] += y + P) %= P;
}
inline long long que(int x) {
  long long res = 0;
  for (; x; x -= x & -x) (res += c[x]) %= P;
  return res;
}
inline long long query(int l, int r) {
  if (r < l) return 0;
  return (que(r) - que(l - 1) + P) % P;
}
}  // namespace MUL
inline int getpos(int ql, int qr) {
  int l = ql, r = qr, mid, res;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (SUM::query(ql, mid) >= SUM::query(mid + 1, qr))
      res = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return res;
}
void solve(int l, int r) {
  if (l == r) return (void)(print(0));
  int pos = getpos(l, r);
  long long res = 0;
  res = (-MUL::query(l, pos) +
         (SUM::query(l, pos) % P) * (long long)abs(a[pos] - pos) % P + P) %
        P;
  res = (res + MUL::query(pos, r) -
         SUM::query(pos, r) % P * (a[pos] - pos) % P + P) %
        P;
  print(res);
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) {
    w[i] = read(), SUM::add(i, w[i]), MUL::add(i, w[i] * (a[i] - i));
  }
  while (q--) {
    int x = read(), y = read();
    if (x < 0) {
      x = -x, SUM::add(x, -w[x]), MUL::add(x, -w[x] * (a[x] - x));
      w[x] = y, SUM::add(x, y), MUL::add(x, y * (a[x] - x));
    } else
      solve(x, y);
  }
  Ot();
  return 0;
}