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>
const int N = 200005, mod = 1000000007;
int n, q, a[N], b[N], w[N];
void up(long long &x, int y) { x += y - mod, x += x >> 63 & mod; }
void up(long long &x, int y, int z) { x = (x + (long long)y * z) % mod; }
long long bit0[N], bit1[N];
void modify(int x, int d) {
  int y = (long long)(d + mod) * (b[x] + mod) % mod;
  for (; x <= n; x += x & -x) bit0[x] += d, up(bit1[x], y);
}
long long query(long long *bit, int l, int r) {
  long long ans = 0;
  for (; r; r &= r - 1) ans += bit[r];
  for (--l; l; l &= l - 1) ans -= bit[l];
  return ans;
}
int query(int l, int r) {
  int L = l, R = r, ans = -1;
  long long sum = query(bit0, l, r);
  while (L <= R) {
    int mid = L + R >> 1;
    if (2 * query(bit0, l, mid) >= sum)
      ans = mid, R = mid - 1;
    else
      L = mid + 1;
  }
  int x = (b[ans] + mod) % mod;
  int left = query(bit0, l, ans - 1) % mod;
  int right = query(bit0, ans + 1, r) % mod;
  int _left = query(bit1, l, ans - 1) % mod;
  int _right = query(bit1, ans + 1, r) % mod;
  return (mod - _left + (long long)x * left + (long long)(mod - x) * right +
          _right) %
         mod;
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n >> q;
  for (int i = 1; i <= n; ++i) std::cin >> a[i], b[i] = a[i] - i;
  for (int i = 1; i <= n; ++i) std::cin >> w[i];
  for (int i = 1; i <= n; ++i)
    bit0[i] = w[i], bit1[i] = (long long)w[i] * (a[i] - i + mod) % mod;
  for (int i = 1; i <= n; ++i) {
    int j = i + (i & -i);
    if (j <= n) bit0[j] += bit0[i], up(bit1[j], bit1[i]);
  }
  while (q--) {
    int x, y;
    std::cin >> x >> y;
    if (x < 0)
      modify(-x, y - w[-x]), w[-x] = y;
    else
      std::cout << query(x, y) << '\n';
  }
  return 0;
}