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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void wri(long long a) {
  write(a);
  putchar(' ');
}
inline void writeln(long long a) {
  write(a);
  puts("");
}
const long long N = 1000005, mod = 1e9 + 7;
long long n, m, tree[N], sum[N], a[N], w[N];
inline void push_up(long long nod) {
  tree[nod] = (tree[nod << 1] + tree[nod << 1 | 1]) % mod;
  sum[nod] = sum[nod << 1] + sum[nod << 1 | 1];
}
void build(long long l, long long r, long long nod) {
  if (l == r) {
    tree[nod] = a[l] * w[l] % mod;
    sum[nod] = w[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, nod << 1);
  build(mid + 1, r, nod << 1 | 1);
  push_up(nod);
}
void insert(long long l, long long r, long long pos, long long nod) {
  if (l == r) {
    tree[nod] = a[l] * w[l] % mod;
    sum[nod] = w[l];
    return;
  }
  long long mid = (l + r) >> 1;
  if (pos <= mid)
    insert(l, mid, pos, nod << 1);
  else
    insert(mid + 1, r, pos, nod << 1 | 1);
  push_up(nod);
}
long long find(long long l, long long r, long long dq, long long nod) {
  if (sum[nod] < dq) return 0;
  if (l == r) return l;
  long long mid = (l + r) >> 1;
  if (sum[nod << 1] >= dq)
    return find(l, mid, dq, nod << 1);
  else
    return find(mid + 1, r, dq - sum[nod << 1], nod << 1 | 1);
}
long long ask(long long l, long long r, long long i, long long j,
              long long nod) {
  if (i > j) return 0;
  if (l == i && r == j) return sum[nod];
  long long mid = (l + r) >> 1;
  if (j <= mid)
    return ask(l, mid, i, j, nod << 1);
  else if (i > mid)
    return ask(mid + 1, r, i, j, nod << 1 | 1);
  else
    return ask(l, mid, i, mid, nod << 1) +
           ask(mid + 1, r, mid + 1, j, nod << 1 | 1);
}
long long Ask(long long l, long long r, long long i, long long j,
              long long nod) {
  if (i > j) return 0;
  if (l == i && r == j) return tree[nod];
  long long mid = (l + r) >> 1;
  if (j <= mid)
    return Ask(l, mid, i, j, nod << 1);
  else if (i > mid)
    return Ask(mid + 1, r, i, j, nod << 1 | 1);
  else
    return (Ask(l, mid, i, mid, nod << 1) +
            Ask(mid + 1, r, mid + 1, j, nod << 1 | 1)) %
           mod;
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i <= n; i++) a[i] = read() - i;
  for (long long i = 1; i <= n; i++) w[i] = read();
  build(1, n, 1);
  for (long long i = 1; i <= m; i++) {
    long long x = read(), y = read();
    if (x > 0) {
      long long t1 = ask(1, n, 1, x - 1, 1), t2 = ask(1, n, 1, y, 1);
      long long t = find(1, n, (t2 + t1 + 1) / 2, 1);
      writeln(
          (((ask(1, n, x, t, 1) % mod * a[t] - Ask(1, n, x, t, 1)) +
            (Ask(1, n, t + 1, y, 1) - ask(1, n, t + 1, y, 1) % mod * a[t])) %
               mod +
           mod) %
          mod);
    } else {
      x = -x;
      w[x] = y;
      insert(1, n, x, 1);
    }
  }
}