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 M = 2e5 + 10;
const long long MOD = 1e9 + 7;
int n, Q;
long long A[M], W[M];
struct bit_tree {
  long long bit[M], Mod;
  void init(long long mod) { Mod = mod; }
  void Add(long long &a, long long b) {
    a += b;
    if (Mod != -1) {
      a %= Mod;
      if (a < 0) a += Mod;
    }
  }
  void add(int a, long long v) {
    if (!a) a = 1;
    for (; a <= n; a += ((a) & (-(a)))) Add(bit[a], v);
  }
  long long query(int a) {
    long long res = 0;
    for (; a > 0; a -= ((a) & (-(a)))) Add(res, bit[a]);
    return res;
  }
} b1, b2;
void change_pos(long long p, long long v) {
  long long res = v - W[p];
  b1.add(p, res);
  b2.add(p, A[p] * res % MOD);
  W[p] += res;
}
void solve_area(long long le, long long ri) {
  long long sum = b1.query(le - 1);
  sum += ((b1.query(ri) - sum + 1) >> 1);
  int l = le, r = ri, mid;
  while (l < r) {
    mid = l + r >> 1;
    if (b1.query(mid) >= sum)
      r = mid;
    else
      l = mid + 1;
  }
  long long all = b2.query(ri) + b2.query(le - 1) - (2ll * b2.query(l) % MOD);
  all %= MOD;
  if (all < 0) all += MOD;
  sum = b1.query(l);
  long long bef = (sum - b1.query(le - 1)) % MOD + sum - b1.query(ri);
  bef %= MOD;
  if (bef < 0) bef += MOD;
  printf("%I64d\n", (all + bef * A[l] % MOD) % MOD);
}
long long a, b;
int main() {
  b1.init(-1);
  b2.init(MOD);
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &A[i]);
    A[i] -= i;
  }
  for (int i = 1; i <= n; i++) scanf("%I64d", &W[i]);
  for (int i = 1; i <= n; i++) b1.add(i, W[i]), b2.add(i, A[i] * W[i] % MOD);
  while (Q--) {
    scanf("%I64d%I64d", &a, &b);
    if (a < 0)
      change_pos(-a, b);
    else
      solve_area(a, b);
  }
  return 0;
}