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;
struct Node {
  int size, xL, xR;
  long long sum, costTL, costTR;
  Node() { size = xL = xR = sum = costTL = costTR = 0; }
  Node(long long x, long long w) {
    xL = xR = x;
    size = 1;
    sum = w;
    costTL = costTR = 0;
  }
  void updateN(long long w) { sum = w; }
};
const int MOD = 1e9 + 7;
Node join(Node p1, Node p2) {
  Node r;
  r.xL = p1.xL, r.xR = p2.xR;
  r.sum = p1.sum + p2.sum;
  r.size = p1.size + p2.size;
  r.costTL = p1.costTL + p2.costTL +
             ((p2.xL - p1.xL - p1.size) % MOD) * (p2.sum % MOD);
  r.costTL %= MOD;
  r.costTR = p1.costTR + p2.costTR +
             ((p2.xR - p1.xR - p2.size) % MOD) * (p1.sum % MOD);
  r.costTR %= MOD;
  return r;
}
int N;
int A[200010], W[200010];
Node st[800010];
void create(int n = 1, int l = 0, int r = N) {
  if (l == r)
    st[n] = Node(A[l], W[l]);
  else {
    int m = (l + r) / 2;
    create(2 * n, l, m);
    create(2 * n + 1, m + 1, r);
    st[n] = join(st[2 * n], st[2 * n + 1]);
  }
}
Node query(int x, int y, int n = 1, int l = 0, int r = N) {
  if (x <= l && r <= y)
    return st[n];
  else {
    int m = (l + r) / 2;
    if (y <= m) return query(x, y, 2 * n, l, m);
    if (x > m) return query(x, y, 2 * n + 1, m + 1, r);
    return join(query(x, y, 2 * n, l, m), query(x, y, 2 * n + 1, m + 1, r));
  }
}
void update(int x, int v, int n = 1, int l = 0, int r = N) {
  if (l == r)
    st[n].updateN(v);
  else {
    int m = (l + r) / 2;
    if (x <= m)
      update(x, v, 2 * n, l, m);
    else
      update(x, v, 2 * n + 1, m + 1, r);
    st[n] = join(st[2 * n], st[2 * n + 1]);
  }
}
int getMid(long long v, int n = 1, int l = 0, int r = N) {
  if (l == r) return l;
  int m = (l + r) / 2;
  if (st[2 * n].sum < v) {
    v -= st[2 * n].sum;
    return getMid(v, 2 * n + 1, m + 1, r);
  }
  return getMid(v, 2 * n, l, m);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  for (int i = 0; i < n; i++) scanf("%d", &W[i]);
  N = n - 1;
  create();
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0)
      update(-x - 1, y);
    else {
      x--, y--;
      if (x == y) {
        puts("0");
        continue;
      }
      long long ms = query(x, y).sum / 2;
      if (x) ms += query(0, x - 1).sum;
      int m = getMid(ms);
      if (m >= y) m--;
      Node t1 = query(x, m), t2 = query(m + 1, y);
      long long ans = t1.costTR + t2.costTL;
      ans += ((t2.xL - t1.xR - 1) % MOD) * (min(t1.sum, t2.sum) % MOD);
      printf("%I64d\n", ans % MOD);
    }
  }
}