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 MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {-1, -2, 1, -2, 2, -1, 2, 1}, ddy[] = {-2, -1, -2, 1, -1, 2, 1, 2};
long long a[200001], b[200001];
long long tree1[524288], tree2[524288];
void init(int node, int S, int E) {
  if (S == E) {
    tree1[node] = (a[S] * b[S]) % MOD;
    tree2[node] = b[S];
    return;
  }
  init(2 * node, S, (S + E) / 2);
  init(2 * node + 1, (S + E) / 2 + 1, E);
  tree1[node] = (tree1[node * 2] + tree1[node * 2 + 1]) % MOD;
  tree2[node] = tree2[node * 2] + tree2[node * 2 + 1];
}
void upd(int node, int S, int E, int k) {
  if (S == E) {
    tree1[node] = (a[S] * b[S]) % MOD;
    tree2[node] = b[S];
    return;
  }
  if (k <= (S + E) / 2)
    upd(node * 2, S, (S + E) / 2, k);
  else
    upd(node * 2 + 1, (S + E) / 2 + 1, E, k);
  tree1[node] = (tree1[node * 2] + tree1[node * 2 + 1]) % MOD;
  tree2[node] = tree2[node * 2] + tree2[node * 2 + 1];
}
pair<long long, long long> operator+(const pair<long long, long long> &p1,
                                     const pair<long long, long long> &p2) {
  return make_pair((p1.first + p2.first) % MOD, p1.second + p2.second);
}
pair<long long, long long> find(int node, int S, int E, int i, int j) {
  if (i > E || j < S) return {0, 0};
  if (i <= S && j >= E) return {tree1[node], tree2[node]};
  return find(node * 2, S, (S + E) / 2, i, j) +
         find(node * 2 + 1, (S + E) / 2 + 1, E, i, j);
}
int findK(int node, int S, int E, long long k) {
  if (S == E) return S;
  if (k <= tree2[node * 2]) return findK(node * 2, S, (S + E) / 2, k);
  return findK(node * 2 + 1, (S + E) / 2 + 1, E, k - tree2[node * 2]);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int(i) = (1); (i) <= (n); (i) += (1)) {
    scanf("%lld", a + i);
    a[i] -= i;
  }
  for (int(i) = (1); (i) <= (n); (i) += (1)) scanf("%lld", b + i);
  init(1, 1, n);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > 0) {
      long long k = (find(1, 1, n, x, y).second + 1) / 2;
      int i = findK(1, 1, n, k + find(1, 1, n, 1, x - 1).second);
      pair<long long, long long> s1 = find(1, 1, n, x, i - 1),
                                 s2 = find(1, 1, n, i + 1, y);
      long long ans =
          ((s1.second * a[i] - s1.first) + (s2.first - s2.second * a[i])) % MOD;
      if (ans < 0) ans += MOD;
      printf("%lld\n", ans);
    } else {
      x = -x;
      b[x] = y;
      upd(1, 1, n, x);
    }
  }
}