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 = 200005;
const int MOD = 1000000007;
int N, Q;
long long A[MAXN];
long long W[MAXN];
long long BIT[3][MAXN];
long long get(int t, int idx) {
  long long ret = 0;
  for (; idx > 0; idx -= idx & -idx) {
    ret += BIT[t][idx];
    if (t != 0) ret %= MOD;
  }
  return ret;
}
void add(int t, int idx, long long val) {
  for (; idx < MAXN; idx += idx & -idx) {
    BIT[t][idx] += val;
    if (t != 0) BIT[t][idx] %= MOD;
  }
}
int get_median(int left, int right) {
  long long pre_left = get(0, left - 1);
  long long pre_right = get(0, right);
  long long sum = pre_right - pre_left;
  int lo = left, hi = right;
  while (lo < hi) {
    int mid = lo + (hi - lo) / 2;
    if (2 * (get(0, mid) - pre_left) >= sum)
      hi = mid;
    else
      lo = mid + 1;
  }
  return lo;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) cin >> A[i];
  for (int i = 1; i <= N; i++) {
    cin >> W[i];
    add(0, i, W[i]);
    add(1, i, W[i] * i % MOD);
    add(2, i, A[i] * W[i] % MOD);
  }
  for (int i = 0, x, y; i < Q; i++) {
    cin >> x >> y;
    if (x < 0) {
      x *= -1;
      add(0, x, y - W[x]);
      add(1, x, x * (y - W[x]) % MOD);
      add(2, x, A[x] * (y - W[x]) % MOD);
      W[x] = y;
    } else {
      int mid = get_median(x, y);
      long long left_sum[3] = {get(0, x - 1), get(1, x - 1), get(2, x - 1)};
      long long mid_sum[3] = {get(0, mid), get(1, mid), get(2, mid)};
      long long right_sum[3] = {get(0, y), get(1, y), get(2, y)};
      long long ans = (mid_sum[1] - left_sum[1]) % MOD;
      ans = (ans + ((A[mid] - mid) % MOD) * ((mid_sum[0] - left_sum[0]) % MOD) %
                       MOD) %
            MOD;
      ans = (ans - mid_sum[2] + left_sum[2]) % MOD;
      ans = (ans - right_sum[1] + mid_sum[1]) % MOD;
      ans = (ans - ((A[mid] - mid) % MOD) *
                       ((right_sum[0] - mid_sum[0]) % MOD) % MOD) %
            MOD;
      ans = (ans + right_sum[2] - mid_sum[2]) % MOD;
      if (ans < 0) ans += MOD;
      cout << ans << "\n";
    }
  }
  return 0;
}