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 long long N = 2e5 + 5;
const long long INF = 1e9 + 7;
const long long mod = 1e9 + 7;
struct data {
  long long a, b, c;
} tree[N << 2 | 1];
long long n, Q, t, a[N], w[N];
data operator+(const data &a, const data &b) {
  data c;
  c.a = a.a + b.a;
  c.b = (a.b + b.b) % mod;
  c.c = (a.c + b.c) % mod;
  return c;
}
inline void init(long long k, long long l, long long r) {
  if (l == r)
    tree[k] = {w[l], w[l] * (l - a[l]) % mod, w[l] * (a[l] - l) % mod};
  else {
    long long mid = (l + r) >> 1;
    init(k << 1, l, mid);
    init(k << 1 | 1, mid + 1, r);
    tree[k] = tree[k << 1] + tree[k << 1 | 1];
  }
}
inline void update(long long k, long long l, long long r, long long pos) {
  if (l > pos || r < pos) return;
  if (l == r && l == pos)
    tree[k] = {w[l], w[l] * (l - a[l]) % mod, w[l] * (a[l] - l) % mod};
  else {
    long long mid = (l + r) >> 1;
    update(k << 1, l, mid, pos);
    update(k << 1 | 1, mid + 1, r, pos);
    tree[k] = tree[k << 1] + tree[k << 1 | 1];
  }
}
inline data query(long long k, long long l, long long r, long long L,
                  long long R) {
  if (l > R || r < L) return {0, 0, 0};
  if (L <= l && r <= R)
    return tree[k];
  else {
    long long mid = (l + r) >> 1;
    return query(k << 1, l, mid, L, R) + query(k << 1 | 1, mid + 1, r, L, R);
  }
}
inline long long sum(const long long &l, const long long &r,
                     const long long &num) {
  if (num == 1) return query(1, 1, n, l, r).a;
  if (num == 2) return query(1, 1, n, l, r).b;
  if (num == 3) return query(1, 1, n, l, r).c;
}
inline long long Find(long long k, long long l, long long r, long long val) {
  if (l == r) return l;
  long long mid = (l + r) >> 1;
  if (t % 2 == 0) {
    if (tree[k << 1].a > val) return Find(k << 1, l, mid, val);
    return Find(k << 1 | 1, mid + 1, r, val - tree[k << 1].a);
  } else {
    if (tree[k << 1].a >= val) return Find(k << 1, l, mid, val);
    return Find(k << 1 | 1, mid + 1, r, val - tree[k << 1].a);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%lld%lld", &n, &Q);
  for (long long i = 1; i <= n; i++) scanf("%lld", a + i);
  for (long long i = 1; i <= n; i++) scanf("%lld", w + i);
  init(1, 1, n);
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    if (x < 0) {
      x = -x;
      w[x] = y;
      update(1, 1, n, x);
    } else {
      long long k = 0;
      t = sum(x, y, 1);
      long long ss = (sum(x, y, 1) + 1) / 2;
      k = Find(1, 1, n, ss + sum(1, x - 1, 1));
      long long s = 0;
      s = (s + (a[k] * (sum(x, k - 1, 1) % mod) % mod) -
           (k * (sum(x, k - 1, 1) % mod) % mod) + mod) %
          mod;
      s = (s + (sum(x, k - 1, 2) % mod) + mod) % mod;
      s = (s - (a[k] * (sum(k, y, 1) % mod) % mod) +
           (k * (sum(k, y, 1) % mod) % mod) + mod) %
          mod;
      s = (s + (sum(k, y, 3) % mod) + mod) % mod;
      printf("%lld\n", s);
    }
  }
  return 0;
}