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>
#pragma optimise("O3")
using namespace std;
mt19937 rnumber((long long)new char);
typedef struct Treap* Arbore;
Arbore NIL;
struct Treap {
  int move_st, move_dr;
  int poz, poz_min, poz_max, w, g;
  int prio;
  long long suma;
  Arbore st, dr;
  Treap(int g, int poz)
      : move_st(0),
        move_dr(0),
        poz(poz),
        poz_min(poz),
        poz_max(poz),
        suma(g),
        w(g),
        prio(rnumber()),
        st(NIL),
        dr(NIL),
        g(1) {}
  inline void recalc() {
    g = 1 + st->g + dr->g;
    suma = w + st->suma + dr->suma;
    poz_min = min(poz, st->poz_min);
    poz_max = max(poz, dr->poz_max);
    move_st =
        (1LL * (poz - poz_min - st->g) * w +
         1LL * (dr->poz_min - poz_min - st->g - 1) * (dr->suma % 1000000007) +
         st->move_st + dr->move_st) %
        1000000007;
    move_dr =
        (1LL * (poz_max - poz - dr->g) * w +
         1LL * (poz_max - st->poz_max - dr->g - 1) * (st->suma % 1000000007) +
         st->move_dr + dr->move_dr) %
        1000000007;
  }
};
Arbore join(Arbore a, Arbore b) {
  if (a == NIL) return b;
  if (b == NIL) return a;
  if (a->prio >= b->prio) {
    a->dr = join(a->dr, b);
    a->recalc();
    return a;
  }
  b->st = join(a, b->st);
  b->recalc();
  return b;
}
pair<Arbore, Arbore> split_by_number(Arbore a, int nr) {
  if (a == NIL) return {NIL, NIL};
  if (a->st->g >= nr) {
    pair<Arbore, Arbore> s = split_by_number(a->st, nr);
    a->st = NIL;
    a->recalc();
    return {s.first, join(s.second, a)};
  }
  pair<Arbore, Arbore> s = split_by_number(a->dr, nr - a->st->g - 1);
  a->dr = NIL;
  a->recalc();
  return {join(a, s.first), s.second};
}
pair<Arbore, Arbore> split_by_value(Arbore a, long long nr) {
  if (a == NIL) return {NIL, NIL};
  if (a->st->suma >= nr) {
    pair<Arbore, Arbore> s = split_by_value(a->st, nr);
    a->st = NIL;
    a->recalc();
    return {s.first, join(s.second, a)};
  }
  pair<Arbore, Arbore> s = split_by_value(a->dr, nr - a->st->suma - a->w);
  a->dr = NIL;
  a->recalc();
  return {join(a, s.first), s.second};
}
Arbore update(Arbore a, int poz, int w) {
  pair<Arbore, Arbore> s1 = split_by_number(a, poz);
  pair<Arbore, Arbore> s2 = split_by_number(s1.first, poz - 1);
  s2.second->w = w;
  s2.second->recalc();
  return join(s2.first, join(s2.second, s1.second));
}
inline Arbore query(Arbore a, int st, int dr) {
  pair<Arbore, Arbore> s1 = split_by_number(a, dr);
  pair<Arbore, Arbore> s2 = split_by_number(s1.first, st - 1);
  Arbore partea_st = s2.first, partea_dr = s1.second;
  a = s2.second;
  long long w = a->suma / 2;
  pair<Arbore, Arbore> s = split_by_value(a, w);
  long long ans = s.first->move_dr + s.second->move_st;
  if (s.first == NIL || s.second == NIL)
    ;
  else if (s.first->suma <= s.second->suma)
    ans += (s.second->poz_min - s.first->poz_max - 1) *
           (s.first->suma % 1000000007);
  else
    ans += (s.second->poz_min - s.first->poz_max - 1) *
           (s.second->suma % 1000000007);
  printf("%d\n", ans % 1000000007);
  return join(join(partea_st, s.first), join(s.second, partea_dr));
}
int main() {
  NIL = new Treap(0, 0);
  NIL->g = 0;
  NIL->poz_min = 1e9, NIL->poz_max = -1e9;
  Arbore k = NIL;
  int n, q;
  scanf("%d%d", &n, &q);
  vector<int> a(n), w(n);
  for (auto& i : a) scanf("%d", &i);
  for (auto& i : w) scanf("%d", &i);
  for (int i = 0; i < n; i++) k = join(k, new Treap(w[i], a[i]));
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a < 0)
      k = update(k, -a, b);
    else
      k = query(k, a, b);
  }
  return 0;
}