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 = 200050;
long long w[N], sum[N], a[N], weg[N], pos[N];
long long ck(long long x, long long mod) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
struct BIT {
  long long sum[N], mod;
  void init(long long m) { mod = m; }
  BIT() {
    for (int i = 0; i < N; i++) sum[i] = 0;
  }
  void Set(int i, long long f) {
    f = ck(f, mod);
    for (; i < N; i += i & -i) sum[i] += f, sum[i] = ck(sum[i], mod);
  }
  long long Get(int i) {
    long long ans = 0;
    for (; i; i -= i & -i) ans += sum[i], ans = ck(ans, mod);
    return ans;
  }
  long long Get(int l, int r) { return ck(Get(r) - Get(l - 1), mod); }
} WTree, STree;
int main() {
  int n, q, l, r, i, j, k;
  scanf("%i %i", &n, &q);
  WTree.init(3e18);
  STree.init(1e9 + 7);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]), a[i] -= i - n;
  for (i = 1; i <= n; i++) scanf("%lld", &w[i]), WTree.Set(i, w[i]);
  for (i = 1; i <= n; i++) STree.Set(i, a[i] * w[i]);
  int mod = 1e9 + 7;
  while (q--) {
    scanf("%i %i", &l, &r);
    if (l < 0) {
      l = -l;
      WTree.Set(l, r - w[l]);
      STree.Set(l, -a[l] * w[l]);
      w[l] = r;
      STree.Set(l, a[l] * w[l]);
    } else {
      int top = r, bot = l, mid;
      long long all = WTree.Get(l, r);
      while (top >= bot) {
        mid = top + bot >> 1;
        long long lo = WTree.Get(l, mid - 1);
        long long hi = WTree.Get(mid + 1, r);
        if (lo > all / 2)
          top = mid - 1;
        else if (hi > all / 2)
          bot = mid + 1;
        else
          break;
      }
      long long fin = a[mid];
      long long lo = STree.Get(l, mid - 1), hi = STree.Get(mid + 1, r);
      long long ow = WTree.Get(l, mid - 1) % mod,
                up = WTree.Get(mid + 1, r) % mod;
      long long ans = (fin * ow % mod - lo + mod) % mod;
      ans += (hi - fin * up % mod + mod) % mod;
      ans %= mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}