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]. 
#include <bits/stdc++.h>
const int N = 2e5 + 5;
const int mod = 1000000007;
using namespace std;
template <class T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  static char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
inline void write(long long x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n, Q;
long long a[N], w[N], b[N];
struct Tree {
  int l, r;
  long long sum, s;
} tree[4 * N];
inline void pushup(int now) {
  tree[now].sum = tree[now << 1].sum + tree[now << 1 | 1].sum;
  tree[now].s = (tree[now << 1].s + tree[now << 1 | 1].s) % mod;
}
inline void build(int now, int l, int r) {
  tree[now].l = l, tree[now].r = r;
  if (l == r) {
    tree[now].sum = w[l];
    tree[now].s = b[l] * w[l] % mod;
    return;
  }
  int m = (l + r) >> 1;
  build(now << 1, l, m);
  build(now << 1 | 1, m + 1, r);
  pushup(now);
}
inline long long query(int now, int l, int r) {
  if (l <= tree[now].l && tree[now].r <= r) return tree[now].sum;
  int m = (tree[now].l + tree[now].r) >> 1;
  long long ans = 0;
  if (l <= m) ans += query(now << 1, l, r);
  if (r > m) ans += query(now << 1 | 1, l, r);
  return ans;
}
inline void update(int now, int pos, int x) {
  if (tree[now].l == tree[now].r) {
    tree[now].sum = x;
    tree[now].s = b[pos] * x % mod;
    return;
  }
  int m = (tree[now].l + tree[now].r) >> 1;
  if (pos <= m)
    update(now << 1, pos, x);
  else
    update(now << 1 | 1, pos, x);
  pushup(now);
}
inline long long ask(int now, int l, int r) {
  if (l <= tree[now].l && tree[now].r <= r) return tree[now].s;
  int m = (tree[now].l + tree[now].r) >> 1;
  long long ans = 0;
  if (l <= m) ans = (ans + ask(now << 1, l, r)) % mod;
  if (r > m) ans = (ans + ask(now << 1 | 1, l, r)) % mod;
  return ans;
}
inline void Solve(int l, int r) {
  int L = l, R = r, ans = l;
  while (L <= R) {
    int mid = (L + R) >> 1;
    long long QL = query(1, l, mid), QR = query(1, mid + 1, r);
    if (QL >= QR)
      ans = mid, R = mid - 1;
    else
      L = mid + 1;
  }
  long long t1 = (query(1, l, ans - 1) % mod * (a[ans] - ans) % mod -
                  ask(1, l, ans - 1) + mod) %
                 mod;
  long long t2 = (query(1, ans + 1, r) % mod * (ans - a[ans] + mod) % mod +
                  ask(1, ans + 1, r) % mod) %
                 mod;
  write((t1 + t2) % mod);
  putchar('\n');
}
int main() {
  read(n), read(Q);
  for (register int i = 1; i <= n; i++) read(a[i]), b[i] = a[i] - i;
  for (register int i = 1; i <= n; i++) read(w[i]);
  build(1, 1, n);
  int l, r;
  while (Q--) {
    read(l), read(r);
    if (l < 0) update(1, -l, r);
    if (l > 0) Solve(l, r);
  }
  return 0;
}
