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 maxn = 1e6 + 500;
const long long inf = 1e18 + 800;
const long long mod = 1e9 + 7;
long long a[maxn];
long long w[maxn];
long long n;
long long fenw[maxn];
long long fenc[maxn];
void update_fenw(long long p, long long v) {
  for (p++; p < maxn; p += (p & (-p))) {
    fenw[p] += v;
  }
}
long long sum_fenw(long long p) {
  long long ans = 0;
  for (p++; p; p -= (p & (-p))) {
    ans += fenw[p];
  }
  return ans;
}
void update_fenc(long long p, long long v) {
  for (p++; p < maxn; p += (p & (-p))) {
    fenc[p] += v;
    fenc[p] %= mod;
  }
}
long long sum_fenc(long long p) {
  long long ans = 0;
  for (p++; p; p -= (p & (-p))) {
    ans += fenc[p];
    ans %= mod;
  }
  return ans;
}
long long find_best(long long v) {
  long long p = 0;
  for (long long i = 18; i >= 0; i--) {
    if (fenw[p + (1 << i)] <= v) {
      p += (1 << i);
      v -= fenw[p];
    }
  }
  return min(n - 2, p);
}
void update(long long id, long long val) {
  update_fenw(id, val - w[id]);
  update_fenc(id, ((val - w[id]) * a[id]) % mod);
  w[id] = val;
}
long long find_res(long long l, long long r, long long y) {
  if (y < l || y > r) {
    return 0;
  }
  long long ans = 0;
  ans -= sum_fenc(y - 1) - sum_fenc(l - 1);
  ans += sum_fenc(r) - sum_fenc(y);
  ans += a[y] * (sum_fenw(y - 1) - sum_fenw(l - 1));
  ans -= a[y] * (sum_fenw(r) - sum_fenw(y));
  ans %= mod;
  ans += mod;
  ans %= mod;
  return ans;
}
long long find_ans(long long l, long long r) {
  long long a = sum_fenw(r);
  long long b = sum_fenw(l - 1);
  long long v = b + (a - b) / 2;
  long long p = find_best(v);
  long long H = 2 * sum_fenw(p) - sum_fenw(l - 1) - sum_fenw(r);
  if (H < 0 && p < r) p++;
  return find_res(l, r, p);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = a[i] - i;
  }
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    update(i, x);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      long long id = -x;
      id--;
      update(id, y);
    } else {
      long long l = x - 1;
      long long r = y - 1;
      long long ans = find_ans(l, r);
      cout << ans << endl;
    }
  }
}