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;
#pragma warning(disable : 4996)
const long long mod = 1000000007;
class BIT {
 public:
  vector<long long> bit;
  int size_ = 0;
  long long MD;
  void init(int sz) {
    size_ = sz + 2;
    bit.resize(size_ + 2, 0);
  }
  void add(int pos, long long x) {
    pos++;
    x %= MD;
    while (pos <= size_) {
      bit[pos] += x;
      bit[pos] %= MD;
      pos += (pos & -pos);
    }
  }
  long long sum(int pos) {
    long long x = 0;
    pos++;
    while (pos >= 1) {
      x += bit[pos];
      x %= MD;
      pos -= (pos & -pos);
    }
    return x % MD;
  }
};
long long N, Q, A[1 << 18], W[1 << 18];
BIT Z1, Z2;
int main() {
  scanf("%lld%lld", &N, &Q);
  Z1.init(N + 2);
  Z2.init(N + 2);
  Z1.MD = (1LL << 60);
  Z2.MD = mod;
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &A[i]);
    A[i] -= i;
  }
  for (int i = 1; i <= N; i++) scanf("%lld", &W[i]);
  for (int i = 1; i <= N; i++) {
    Z1.add(i, W[i]);
    Z2.add(i, A[i] * W[i] % mod);
  }
  for (int i = 1; i <= Q; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if (a < 0) {
      a *= -1;
      Z1.add(a, 1LL * b - W[a]);
      Z2.add(a, 1LL * A[a] * (b - W[a]));
      W[a] = b;
    } else {
      int l = a, r = b;
      int cl = a, cr = b + 1, mid, minx = b;
      for (int t = 1; t <= 30; t++) {
        mid = (cl + cr) / 2;
        long long V1 = (Z1.sum(mid) - Z1.sum(a - 1));
        long long V2 = (Z1.sum(b) - Z1.sum(mid));
        if (V1 < V2) {
          cl = mid;
        } else {
          cr = mid;
          minx = min(minx, mid);
        }
      }
      mid = minx;
      long long wl = (Z1.sum(mid) - Z1.sum(l - 1)) % mod,
                wr = (Z1.sum(r) - Z1.sum(mid)) % mod;
      long long xl = (Z2.sum(mid) - Z2.sum(l - 1)) % mod,
                xr = (Z2.sum(r) - Z2.sum(mid)) % mod;
      long long al = 1LL * A[mid] * wl - xl, ar = xr - 1LL * A[mid] * wr;
      al += mod * mod;
      al %= mod;
      ar += mod * mod;
      ar %= mod;
      printf("%lld\n", (al + ar) % mod);
    }
  }
  return 0;
}