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 MAX = 2e5 + 7;
const int MOD = 1e9 + 7;
using namespace std;
template <typename T>
struct BIT {
  const int n, P;
  vector<T> c;
  BIT(int n, int P) : n(n - 1), c(n), P(P){};
  void update(int i, T k) {
    while (i <= n) {
      c[i] += k;
      if (P) c[i] %= P;
      i += i & (-i);
      ;
    }
  }
  T getsum(int i) {
    T res = 0;
    while (i > 0) {
      res += c[i];
      if (P) res %= P;
      i -= i & (-i);
      ;
    }
    return res;
  }
};
int n, q;
long long a[MAX], w[MAX], d[MAX];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> q;
  BIT<long long> t1(n + 1, 0), t2(n + 1, MOD);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    d[i] = (a[i] - i + MOD) % MOD;
  }
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    t1.update(i, w[i]);
    t2.update(i, d[i] * w[i] % MOD);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      long long cg1 = y - (t1.getsum(x) - t1.getsum(x - 1)),
                cg2 =
                    (d[x] * y - (t2.getsum(x) - t2.getsum(x - 1)) + MOD) % MOD;
      t1.update(x, cg1);
      t2.update(x, cg2);
    } else {
      int lt = x, rt = y;
      long long rsum = t1.getsum(y), lsum = t1.getsum(x - 1);
      while (lt < rt) {
        int mi = (lt + rt) >> 1;
        long long cnt = t1.getsum(mi) - lsum;
        if (2 * cnt >= rsum - lsum)
          rt = mi;
        else
          lt = mi + 1;
      }
      long long ans1 = 0, ans2 = 0;
      if (lt > x)
        ans1 = (((t1.getsum(lt - 1) - lsum) % MOD * d[lt] % MOD) -
                ((t2.getsum(lt - 1) - t2.getsum(x - 1) + MOD) % MOD) + MOD) %
               MOD;
      if (lt < y)
        ans2 = (((t2.getsum(y) - t2.getsum(lt) + MOD) % MOD) -
                ((rsum - t1.getsum(lt)) % MOD * d[lt] % MOD) + MOD) %
               MOD;
      cout << (ans1 + ans2) % MOD << '\n';
    }
  }
  return 0;
}