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 p, i, j, k, l, s, n, m, sum, x, y, a[400005], b[400005];
struct node {
  long long sum, val;
} f[400005 * 4];
inline node up(node x, node y) {
  return node{(x.sum + y.sum), (x.val + y.val) % 1000000007};
}
inline void change(long long o, long long l, long long r, long long ll,
                   long long p) {
  if (l == ll && r == ll) {
    f[o].val = a[l] * p % 1000000007;
    f[o].sum = p;
    return;
  }
  long long mid = (l + r) >> 1;
  if (ll <= mid)
    change(o * 2, l, mid, ll, p);
  else
    change(o * 2 + 1, mid + 1, r, ll, p);
  f[o] = up(f[o * 2], f[o * 2 + 1]);
}
inline node ask(long long o, long long l, long long r, long long ll,
                long long rr) {
  if (l == ll && r == rr) return f[o];
  long long mid = (l + r) >> 1;
  if (rr <= mid)
    return ask(o * 2, l, mid, ll, rr);
  else if (ll > mid)
    return ask(o * 2 + 1, mid + 1, r, ll, rr);
  else
    return up(ask(o * 2, l, mid, ll, mid),
              ask(o * 2 + 1, mid + 1, r, mid + 1, rr));
}
inline long long get(long long o, long long l, long long r, long long ll,
                     long long rr) {
  if (p * 2 >= sum) return 0;
  if (l >= ll && r <= rr) {
    if (p * 2 >= sum) return 0;
    if ((p + f[o].sum) * 2 < sum) {
      p += f[o].sum;
      return 0;
    }
    if (l == r) {
      p += f[o].sum;
      return l;
    }
    long long mid = (l + r) >> 1;
    if (2 * (f[o * 2].sum + p) >= sum)
      return get(o * 2, l, mid, ll, rr);
    else {
      p += f[o * 2].sum;
      return get(o * 2 + 1, mid + 1, r, ll, rr);
    }
  }
  long long mid = (l + r) >> 1;
  if (rr <= mid)
    return get(o * 2, l, mid, ll, rr);
  else if (ll > mid)
    return get(o * 2 + 1, mid + 1, r, ll, rr);
  else {
    long long xx = get(o * 2, l, mid, ll, mid);
    long long yy = get(o * 2 + 1, mid + 1, r, mid + 1, rr);
    return max(xx, yy);
  }
}
inline long long ask(long long x, long long l, long long r) {
  node gt = ask(1, 1, n, l, x);
  node gtw = ask(1, 1, n, x, r);
  return (gt.sum % 1000000007 * a[x] % 1000000007 - gt.val + gtw.val -
          gtw.sum % 1000000007 * a[x] % 1000000007) %
         1000000007;
}
int main() {
  scanf("%I64d%I64d", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]), a[i] -= i;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &b[i]);
    change(1, 1, n, i, b[i]);
  }
  for (long long i = 1; i <= m; i++) {
    scanf("%I64d%I64d", &x, &y);
    if (x > 0) {
      sum = ask(1, 1, n, x, y).sum;
      p = 0;
      long long gt = get(1, 1, n, x, y);
      printf("%I64d\n",
             (((ask(gt, x, y)) % 1000000007) + 1000000007) % 1000000007);
    } else {
      x = -x;
      b[x] = y;
      change(1, 1, n, x, b[x]);
    }
  }
}