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 aw[200001], w[200001], p[200001], st[200001], sumst[200001],
    sumaw[200001], sumw[200001], cd, mod = 1e9 + 7, q, x, y;
void update(long long a[], long long x, long long val, int type) {
  while (x <= cd) {
    a[x] += val;
    if (type == 0) a[x] %= mod;
    x += (x & -x);
  }
}
void inp() {
  cin >> cd >> q;
  for (long long i = 1; i <= cd; i++) cin >> p[i];
  for (long long i = 1; i <= cd; i++) cin >> w[i];
  for (long long i = 1; i <= cd; i++) {
    update(sumaw, i, w[i] * p[i], 0);
    aw[i] = w[i] * p[i];
    update(sumw, i, w[i], 1);
    update(sumst, i, w[i] * i, 0);
    st[i] = w[i] * i;
  }
}
long long get(long long a[], long long x, long long type) {
  long long k = 0;
  while (x > 0) {
    k += a[x];
    if (type == 0) k %= mod;
    x -= (x & -x);
  }
  return k;
}
long long getaw(long long l, long long r) {
  if (l > r) return 0;
  return (get(sumaw, r, 0) - get(sumaw, l - 1, 0) + mod * 5) % mod;
}
long long getw(long long l, long long r) {
  if (l > r) return 0;
  return (get(sumw, r, 1) - get(sumw, l - 1, 1));
}
long long getst(long long l, long long r, long long st) {
  if (l > r) return 0;
  return (get(sumst, r, 0) - get(sumst, l - 1, 0) -
          (l - st) * (getw(l, r) % mod) % mod + mod * 10) %
         mod;
}
void do_it() {
  for (long long i = 1; i <= q; i++) {
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      update(sumw, x, y - w[x], 1);
      w[x] = y;
      update(sumaw, x, y * p[x] - aw[x], 0);
      aw[x] = y * p[x];
      update(sumst, x, y * x - st[x], 0);
      st[x] = y * x;
    } else {
      long long l = x, r = y, pos, need = (getw(l, r) + 1) / 2;
      while (r >= l) {
        long long mid = (l + r) / 2;
        if (getw(x, mid) >= need)
          pos = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      long long t = pos;
      pos = p[pos] - (pos - x);
      cout << (-(getaw(x, t) - getw(x, t) % mod * pos % mod - getst(x, t, 0) +
                 mod * 5) +
               (getaw(t + 1, y) - getw(t + 1, y) % mod * pos % mod -
                getst(t + 1, y, t - x + 1)) +
               mod * 1000) %
                  mod
           << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  inp(), do_it();
}