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 mo = 1e9 + 7;
const long long maxn = 2e5 + 100;
long long n, Q, x, y, a[maxn], w[maxn], F[maxn], G[maxn], H[maxn], l, r, mid,
    pos, lim, ans;
long long read() {
  long long tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
int lowbit(int x) { return x & -x; }
void modifyF(long long x, long long y) {
  for (int i = x; i <= n; i += lowbit(i)) F[i] += y;
}
long long queryF(long long x) {
  long long res = 0;
  for (int i = x; i > 0; i -= lowbit(i)) res += F[i];
  return res;
}
void modifyG(long long x, long long y) {
  for (int i = x; i <= n; i += lowbit(i)) G[i] = (G[i] + y) % mo;
}
long long queryG(long long x) {
  long long res = 0;
  for (int i = x; i > 0; i -= lowbit(i)) res = (res + G[i]) % mo;
  return res;
}
void modifyH(long long x, long long y) {
  for (int i = x; i <= n; i += lowbit(i)) H[i] = (H[i] + y) % mo;
}
long long queryH(long long x) {
  long long res = 0;
  for (int i = x; i > 0; i -= lowbit(i)) res = (res + H[i]) % mo;
  return res;
}
int main() {
  n = read();
  Q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) {
    w[i] = read();
    modifyF(i, w[i]);
    modifyG(i, (a[i] * w[i]) % mo);
    modifyH(i, (i * w[i]) % mo);
  }
  while (Q) {
    x = read();
    y = read();
    if (x < 0) {
      x = -x;
      modifyF(x, y - w[x]);
      modifyG(x, a[x] * (y - w[x]) % mo);
      modifyH(x, x * (y - w[x]) % mo);
      w[x] = y;
    } else {
      l = x;
      r = y;
      lim = (queryF(y) - queryF(x - 1)) / 2;
      while (l + 1 < r) {
        mid = (l + r) / 2;
        if (queryF(mid - 1) - queryF(x - 1) <= lim)
          l = mid;
        else
          r = mid - 1;
      }
      if (l + 1 <= y && queryF(l + 1 - 1) - queryF(x - 1) <= lim) l++;
      pos = l;
      ans = 0;
      ans += ((queryF(pos) - queryF(x - 1)) % mo * (a[pos] - pos) % mo -
              (queryG(pos) - queryG(x - 1)) % mo +
              (queryH(pos) - queryH(x - 1)) % mo + mo * 4ll) %
             mo;
      ans += ((queryF(y) - queryF(pos - 1)) % mo * (pos - a[pos]) % mo +
              (queryG(y) - queryG(pos - 1)) % mo -
              (queryH(y) - queryH(pos - 1)) % mo + mo * 4ll) %
             mo;
      ans %= mo;
      cout << ans << "\n";
    }
    Q--;
  }
  return 0;
}