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;
int n, q, x, y, lo[200050];
long long ans;
struct dick {
  long long l, w;
} a[200050], b[200050];
struct data {
  long long l, r, s;
} tr1[4 * 200050], tr2[4 * 200050];
void build(int la, int l1, int r1) {
  tr1[la].l = l1;
  tr1[la].r = r1;
  tr2[la].l = l1;
  tr2[la].r = r1;
  if (l1 == r1) {
    tr1[la].s = a[l1].w;
    lo[l1] = la;
    tr2[la].s = b[l1].w * b[l1].l;
    return;
  }
  build(2 * la, l1, (l1 + r1) / 2);
  build(2 * la + 1, (l1 + r1) / 2 + 1, r1);
  tr1[la].s = tr1[2 * la].s + tr1[2 * la + 1].s;
  tr2[la].s = tr2[2 * la].s + tr2[2 * la + 1].s;
  tr2[la].s %= 1000000007;
}
void ch(int lab) {
  long long mid = lo[lab], mid1 = y - a[lab].w,
            mid2 = (y - b[lab].w) * b[lab].l % 1000000007;
  a[lab].w = b[lab].w = y;
  while (mid) {
    tr1[mid].s += mid1;
    tr2[mid].s += mid2;
    tr2[mid].s %= 1000000007;
    mid >>= 1;
  }
}
long long q1(int la, int l1, int r1) {
  if (l1 <= tr1[la].l && r1 >= tr1[la].r) return tr1[la].s;
  long long dick = 0;
  if (l1 <= tr1[2 * la].r) dick += q1(2 * la, l1, r1);
  if (r1 >= tr1[2 * la + 1].l) dick += q1(2 * la + 1, l1, r1);
  return dick;
}
long long q2(int la, int l1, int r1) {
  if (l1 <= tr2[la].l && r1 >= tr2[la].r) return tr2[la].s % 1000000007;
  long long dick = 0;
  if (l1 <= tr2[2 * la].r) dick += q2(2 * la, l1, r1) % 1000000007;
  if (r1 >= tr2[2 * la + 1].l) dick += q2(2 * la + 1, l1, r1) % 1000000007;
  return dick % 1000000007;
}
long long caul(int l, int r, int k) {
  long long sl = 0, sr = 0;
  if (k > l) sl = (q1(1, l, k - 1) % 1000000007) * b[k].l - q2(1, l, k - 1);
  if (k < r) sr = q2(1, k + 1, r) - (q1(1, k + 1, r) % 1000000007) * b[k].l;
  return ((sl + sr) % 1000000007 + 1000000007) % 1000000007;
}
void deal(int l, int r, long long ll, long long lr, int lm) {
  long long mid = (l + r) / 2, mid1 = 0;
  if (l == r) {
    printf("%lld\n", caul(x, y, l));
    return;
  }
  if (lm == -1) ll = q1(1, x, mid), lr = q1(1, mid + 1, y);
  if (lm != -1) {
    if (lm < mid) {
      mid1 = q1(1, lm + 1, mid);
      ll += mid1, lr -= mid1;
    }
    if (lm > mid) {
      mid1 = q1(1, mid + 1, lm);
      ll -= mid1, lr += mid1;
    }
  }
  if (ll < lr)
    deal(mid + 1, r, ll, lr, mid);
  else
    deal(l, mid, ll, lr, mid);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].l), b[i].l = a[i].l - i;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].w), b[i].w = a[i].w;
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &x, &y);
    if (x < 0)
      ch(-x);
    else
      deal(x, y, 0, 0, -1);
  }
}