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 long long maxn = 2e5 + 10, mod = 1e9 + 7;
pair<long long, long long> p[maxn];
struct Fenwick {
  long long sm[maxn];
  void add(int ind, long long x, bool is) {
    if (is) x %= mod;
    while (ind < maxn) {
      sm[ind] += x;
      if (is) sm[ind] %= mod;
      ind += (ind & -ind);
    }
  }
  long long ask(int ind, bool is) {
    long long ans = 0;
    while (ind > 0) {
      ans += sm[ind];
      if (is) ans %= mod;
      ind -= (ind & -ind);
    }
    return ans;
  }
};
Fenwick fen1, fen2;
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = int(1); i <= int(n); i++) {
    scanf("%lld", &p[i].first);
    p[i].first -= i;
  }
  for (int i = int(1); i <= int(n); i++) {
    scanf("%lld", &p[i].second);
  }
  for (int i = int(1); i <= int(n); i++) {
    fen1.add(i, p[i].first * p[i].second, 1);
    fen2.add(i, p[i].second, 0);
  }
  while (q--) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if (a < 0) {
      a = -a;
      fen1.add(a, -p[a].first * p[a].second, 1);
      fen2.add(a, -p[a].second, 0);
      p[a].second = b;
      fen1.add(a, p[a].first * p[a].second, 1);
      fen2.add(a, p[a].second, 0);
    } else {
      long long bf = fen2.ask(a - 1, 0);
      long long tot = fen2.ask(b, 0) - bf;
      long long l = a, r = b, ans = -1;
      while (l <= r) {
        int mid = (l + r) / 2;
        if ((2 * (fen2.ask(mid, 0) - bf)) >= tot) {
          ans = mid;
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      long long BF = fen1.ask(a - 1, 1);
      long long ANS =
          (p[ans].first * ((2 * (fen2.ask(ans, 0) - bf) - tot) % mod) -
           2 * (fen1.ask(ans, 1) - BF) + (fen1.ask(b, 1) - BF)) %
          mod;
      if (ANS < 0) ANS += mod;
      printf("%lld\n", ANS);
    }
  }
}