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 inf = 0x7fffffff;
const double eps = 1e-10;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch;
  ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch & 15);
    ch = getchar();
  }
  if (f)
    return x;
  else
    return -x;
}
const int N = 2e5 + 10;
const long long mod = 1e9 + 7;
int n, q, w[N], a[N], x, y;
struct BIT1 {
  long long sum[N];
  void add(int i, long long c) {
    for (; i <= n; i += (i & (-i))) sum[i] = sum[i] + c;
  }
  long long query(int i) {
    long long ret = 0;
    for (; i >= 1; i -= (i & (-i))) ret = ret + sum[i];
    return ret;
  }
} bit1;
struct BIT2 {
  long long sum[N];
  void add(int i, long long c) {
    for (; i <= n; i += (i & (-i))) sum[i] = ((sum[i] + c) % mod + mod) % mod;
  }
  long long query(int i) {
    long long ret = 0;
    for (; i >= 1; i -= (i & (-i))) ret = (ret + sum[i]) % mod;
    return ret;
  }
} bit2;
void update(int i, int c) {
  int ls = c - w[i];
  bit1.add(i, 1ll * ls);
  bit2.add(i, 1ll * ls * a[i] % mod);
  w[i] = c;
}
int query(int L, int R) {
  long long ls = bit1.query(x - 1);
  ls = ((bit1.query(y) - ls + 1) >> 1) + ls;
  int l = L, r = R;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (bit1.query(mid) >= ls)
      r = mid;
    else
      l = mid + 1;
  }
  return (int)(((bit2.query(R) + bit2.query(L - 1) - bit2.query(l) -
                 bit2.query(l) + 2 * mod) %
                    mod +
                (bit1.query(l) + bit1.query(l) - bit1.query(L - 1) -
                 bit1.query(R) + 2 * mod) %
                    mod * a[l] % mod) %
               mod);
}
int main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; i++) a[i] = read() - i;
  for (int i = 1; i <= n; i++) w[i] = read();
  for (int i = 1; i <= n; ++i)
    bit1.add(i, 1ll * w[i]), bit2.add(i, 1ll * w[i] * a[i] % mod);
  while (q--) {
    x = read();
    y = read();
    if (x < 0)
      update(-x, y);
    else
      printf("%d\n", query(x, y));
  }
  return 0;
}