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 MAXN = 200010;
const int mod = 1000000007;
void reduce(int& x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
int remod(long long x) {
  x %= mod;
  return x + (x >> 63 & mod);
}
int n;
int A[MAXN], W[MAXN];
long long rt[MAXN];
void mdfc(int x, int v) {
  for (; x <= n; x += x & -x) rt[x] += v;
}
long long qry(int x) {
  long long res = 0;
  for (; x; x &= x - 1) res += rt[x];
  return res;
}
long long qry(int l, int r) { return qry(r) - qry(l - 1); }
int tr[MAXN];
void mdfv(int x, int v) {
  for (v -= mod; x <= n; x += x & -x) reduce(tr[x] += v);
}
int qryv(int x) {
  int res = 0;
  for (; x; x &= x - 1) reduce(res += tr[x] - mod);
  return res;
}
int qryv(int l, int r) {
  int x = qryv(r) - qryv(l - 1);
  return x + (x >> 31 & mod);
}
void add(int x, int v) { mdfc(x, v), mdfv(x, remod(mul(A[x], v))); }
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  int T;
  std::cin >> n >> T;
  for (int i = 1; i <= n; ++i) std::cin >> A[i], A[i] -= i;
  for (int i = 1; i <= n; ++i) std::cin >> W[i], add(i, W[i]);
  while (T-- > 0) {
    int x, y;
    std::cin >> x >> y;
    if (x < 0)
      x = -x, add(x, y - W[x]), W[x] = y;
    else {
      int l = x, r = y, at = -1;
      long long sm = qry(l, r);
      while (l <= r) {
        int mid = l + r >> 1;
        if (qry(x, mid) * 2 >= sm)
          at = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      l = x, r = y;
      int X = A[at];
      long long ans = qry(l, at) % mod * X - qryv(l, at);
      ans += qryv(at + 1, r) - qry(at + 1, r) % mod * X;
      std::cout << remod(ans) << '\n';
    }
  }
  return 0;
}