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 inline read() {
  long long num = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num * neg;
}
const long long maxn = 200010, mod = 1e9 + 7, M = 1e9, xscakioi = 1;
long long n, q, a[maxn], w[maxn];
struct BIT {
  long long sum[maxn];
  void update(long long x, long long k, long long op) {
    for (long long i = x; i <= n; i += i & (-i)) {
      sum[i] = sum[i] + k;
      if (op) sum[i] %= mod;
    }
  }
  long long Query(long long x, long long op) {
    long long ans = 0;
    for (long long i = x; i; i -= i & (-i)) {
      ans = ans + sum[i];
      if (op) ans %= mod;
    }
    return ans;
  }
  long long Q(long long l, long long r, long long op) {
    long long ans = Query(r, op) - Query(l - 1, op);
    if (op) ans = (ans + mod) % mod;
    return ans;
  }
} sum, cnt;
signed main() {
  n = read();
  q = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) w[i] = read();
  for (long long i = 1; i <= n; i++)
    sum.update(i, w[i], 0), cnt.update(i, (a[i] - i) * w[i], 1);
  for (long long i = 1; i <= q; i++) {
    long long l = read(), r = read();
    if (l <= 0) {
      l = -l;
      sum.update(l, -w[l], 0);
      cnt.update(l, -(a[l] - l) * w[l], 1);
      w[l] = r;
      sum.update(l, w[l], 0);
      cnt.update(l, (a[l] - l) * w[l], 1);
    } else {
      if (l == r) {
        cout << 0 << endl;
        continue;
      }
      long long L = l, R = r, ans = 0, res;
      while (L <= R) {
        long long mid = L + R >> 1;
        if (sum.Q(l, mid, 0) >= sum.Q(mid + 1, r, 0))
          res = mid, R = mid - 1;
        else
          L = mid + 1;
      }
      L = res;
      ans = (sum.Q(l, L, 1) % mod * (a[L] - L) % mod - cnt.Q(l, L, 1) % mod +
             mod) %
            mod;
      ans = ((ans + cnt.Q(L, r, 1) % mod) % mod -
             sum.Q(L, r, 1) % mod * (a[L] - L) % mod + mod) %
            mod;
      cout << (ans + mod) % mod << endl;
    }
  }
  return 0;
}