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 maxn = 3e5 + 50;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long a[maxn];
long long w[maxn];
struct node {
  int l, r;
  long long sum, tot;
} tree[maxn << 2];
void push_up(int i) {
  tree[i].sum = tree[i << 1].sum + tree[i << 1 | 1].sum;
  tree[i].tot = (tree[i << 1].tot + tree[i << 1 | 1].tot) % mod;
}
void build(int i, int l, int r) {
  tree[i].l = l;
  tree[i].r = r;
  if (l == r) {
    tree[i].sum = w[l];
    tree[i].tot = (a[l] - 1ll * l) % mod * w[l] % mod;
    return;
  }
  int mid = (tree[i].l + tree[i].r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  push_up(i);
}
void updata(int i, int x, int k) {
  if (tree[i].l == tree[i].r && tree[i].l == x) {
    tree[i].sum = k;
    tree[i].tot = (a[tree[i].l] - 1ll * tree[i].l) % mod * k % mod;
    return;
  }
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (x <= mid)
    updata(i << 1, x, k);
  else
    updata(i << 1 | 1, x, k);
  push_up(i);
}
long long querytot(int i, int x, int y) {
  if (tree[i].l >= x && tree[i].r <= y) return tree[i].tot % mod;
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (y <= mid)
    return querytot(i << 1, x, y);
  else if (x > mid)
    return querytot(i << 1 | 1, x, y);
  else
    return (querytot(i << 1, x, mid) % mod +
            querytot(i << 1 | 1, mid + 1, y) % mod + mod) %
           mod;
}
long long querysum(int i, int x, int y) {
  if (tree[i].l >= x && tree[i].r <= y) return tree[i].sum;
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (y <= mid)
    return querysum(i << 1, x, y);
  else if (x > mid)
    return querysum(i << 1 | 1, x, y);
  else
    return querysum(i << 1, x, mid) + querysum(i << 1 | 1, mid + 1, y);
}
pair<long long, long long> query(int i, int x, int y, long long k) {
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (tree[i].l >= x && tree[i].r <= y) {
    if (tree[i].sum < k) return {tree[i].sum, inf};
    if (tree[i].l == tree[i].r) return {tree[i].sum, tree[i].l};
    if (tree[i << 1].sum >= k)
      return query(i << 1, x, mid, k);
    else
      return query(i << 1 | 1, mid + 1, y, k - tree[i << 1].sum);
  }
  if (y <= mid)
    return query(i << 1, x, y, k);
  else if (x > mid)
    return query(i << 1 | 1, x, y, k);
  else {
    pair<long long, long long> lt = query(i << 1, x, mid, k);
    pair<long long, long long> rt;
    long long sum = querysum(1, x, mid);
    if (lt.second == inf)
      rt = query(i << 1 | 1, mid + 1, y, k - sum);
    else
      return lt;
    return rt;
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]);
  build(1, 1, n);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = abs(x);
      updata(1, x, y);
    } else {
      long long sum = querysum(1, x, y);
      pair<long long, long long> pr = query(1, x, y, (sum + 1) / 2);
      int pos = pr.second;
      long long tot1 = 0, tot2 = 0;
      long long sum1 = 0, sum2 = 0;
      if (x <= pos - 1) {
        sum1 = querysum(1, x, pos - 1);
        tot1 = querytot(1, x, pos - 1);
      }
      tot1 = -tot1;
      if (y >= pos + 1) {
        sum2 = querysum(1, pos + 1, y);
        tot2 = querytot(1, pos + 1, y);
      }
      sum1 %= mod;
      sum2 %= mod;
      long long ans1 = (a[pos] - pos + mod) % mod * sum1 % mod;
      ans1 = (ans1 + tot1 + mod) % mod;
      long long ans2 = (pos - a[pos] + mod) % mod * sum2 % mod;
      ans2 = (ans2 + tot2 + mod) % mod;
      printf("%lld\n", (ans1 + ans2) % mod);
    }
  }
}