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 int N = 200005, mod = 1e9 + 7;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
  return x * f;
}
int n, Q;
int a[N], w[N];
long long c1[N];
int c2[N];
void add(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
void add1(int x, long long d) {
  for (; x <= n; x += x & -x) c1[x] += d;
}
void add2(int x, int d) {
  for (; x <= n; x += x & -x) add(c2[x], d);
}
long long ask1(int x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += c1[x];
  return ans;
}
int ask2(int x) {
  int ans = 0;
  for (; x; x -= x & -x) add(ans, c2[x]);
  return ans;
}
void opt1(int i, int nw) {
  add1(i, -w[i]);
  add2(i, (-1LL * w[i] * a[i] % mod + mod) % mod);
  w[i] = nw;
  add1(i, w[i]);
  add2(i, 1LL * w[i] * a[i] % mod);
}
void opt2(int L, int R) {
  long long t = ask1(L - 1);
  long long sum = ask1(R) - t;
  int le = L, ri = R, mid, ans = R;
  while (le <= ri) {
    mid = (le + ri) >> 1;
    long long v = ask1(mid) - t;
    if (v * 2 >= sum)
      ri = mid - 1, ans = mid;
    else
      le = mid + 1;
  }
  long long v11 = ask2(ans) - ask2(L - 1),
            v12 = -1LL * a[ans] * ((ask1(ans) - t) % mod) % mod;
  long long v21 = ask2(R) - ask2(ans),
            v22 = -1LL * a[ans] * ((ask1(R) - ask1(ans)) % mod) % mod;
  long long Ans = (v21 + v22 - v11 - v12) % mod;
  Ans = (Ans + mod) % mod;
  printf("%I64d\n", Ans);
}
int main() {
  n = read(), Q = read();
  for (int i = 1; i <= n; i++) a[i] = read() - i + n;
  memset(c1, 0, sizeof c1);
  memset(c2, 0, sizeof c2);
  for (int i = 1; i <= n; i++) {
    w[i] = read();
    add1(i, w[i]);
    add2(i, 1LL * w[i] * a[i] % mod);
  }
  while (Q--) {
    int x = read(), y = read();
    if (x < 0)
      opt1(-x, y);
    else
      opt2(x, y);
  }
  return 0;
}