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;
long long n, q, t1[800000];
long long w[800000], a[800000], t2[800000];
void build(int left, int right, int p) {
  if (left == right) {
    t1[p] = w[left];
    t2[p] = w[left] * (left - a[left]);
    t2[p] %= 1000000007;
    return;
  }
  int mid = (left + right) >> 1;
  build(left, mid, p * 2);
  build(mid + 1, right, p * 2 + 1);
  t1[p] = t1[p * 2] + t1[p * 2 + 1];
  t2[p] = t2[p * 2] + t2[p * 2 + 1];
  t2[p] %= 1000000007;
}
void update1(int left, int right, int q, int p, long long c) {
  if (left > q || right < q) {
    return;
  }
  if (left == right) {
    t1[p] += c;
    return;
  }
  int mid = (left + right) >> 1;
  update1(left, mid, q, p * 2, c);
  update1(mid + 1, right, q, p * 2 + 1, c);
  t1[p] = t1[p * 2] + t1[p * 2 + 1];
}
void update2(int left, int right, long long q, int p, long long c) {
  if (left > q || right < q) {
    return;
  }
  if (left == right) {
    t2[p] += c * (q - a[q]);
    t2[p] %= 1000000007;
    return;
  }
  int mid = (left + right) >> 1;
  update2(left, mid, q, p * 2, c);
  update2(mid + 1, right, q, p * 2 + 1, c);
  t2[p] = t2[p * 2] + t2[p * 2 + 1];
  t2[p] %= 1000000007;
}
long long query1(int left, int right, int ql, int qr, int p) {
  if (left >= ql && right <= qr) {
    return t1[p];
  }
  if (left > qr || right < ql) {
    return 0;
  }
  long long mid = (left + right) >> 1;
  return query1(left, mid, ql, qr, p * 2) +
         query1(mid + 1, right, ql, qr, p * 2 + 1);
}
long long query2(int left, int right, int ql, int qr, int p) {
  if (left >= ql && right <= qr) {
    return t2[p];
  }
  if (left > qr || right < ql) {
    return 0;
  }
  int mid = (left + right) >> 1;
  return (query2(left, mid, ql, qr, p * 2) +
          query2(mid + 1, right, ql, qr, p * 2 + 1)) %
         1000000007;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
  }
  build(1, n, 1);
  while (q--) {
    long long x, y;
    scanf("%I64d%I64d", &x, &y);
    if (x < 0) {
      x = -x;
      update1(1, n, x, 1, y - w[x]);
      update2(1, n, x, 1, y - w[x]);
      w[x] = y;
    } else {
      long long l = 1, r = n, res = 0;
      while (l <= r) {
        long long mid = (l + r) >> 1;
        if (query1(1, n, x, mid, 1) < query1(1, n, mid + 1, y, 1)) {
          l = mid + 1;
        } else {
          res = mid;
          r = mid - 1;
        }
      }
      long long ans =
          (query1(1, n, x, res - 1, 1) % 1000000007 * (a[res] - res)) %
              1000000007 +
          query2(1, n, x, res - 1, 1) +
          (query1(1, n, res + 1, y, 1) % 1000000007 * (res - a[res])) %
              1000000007 -
          query2(1, n, res + 1, y, 1);
      while (ans < 0) ans += 1000000007;
      printf("%I64d\n", ans % 1000000007);
    }
  }
  return 0;
}