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 MAXSIZE = 30000020;
int bufpos;
char buf[MAXSIZE];
void init() {
  buf[fread(buf, 1, MAXSIZE, stdin)] = '\0';
  bufpos = 0;
}
int readint() {
  bool isneg;
  int val = 0;
  for (; !isdigit(buf[bufpos]) && buf[bufpos] != '-'; bufpos++)
    ;
  bufpos += (isneg = buf[bufpos] == '-');
  for (; isdigit(buf[bufpos]); bufpos++) val = val * 10 + buf[bufpos] - '0';
  return isneg ? -val : val;
}
char readchar() {
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  return buf[bufpos++];
}
int readstr(char* s) {
  int cur = 0;
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  for (; !isspace(buf[bufpos]); bufpos++) s[cur++] = buf[bufpos];
  s[cur] = '\0';
  return cur;
}
const int maxn = 200002;
const int mod = 1000000007;
struct bit {
  int n;
  long long a[maxn];
  void add(int p, long long v) {
    for (; p <= n; p += p & -p) a[p] += v;
  }
  long long query(int p) {
    long long ans = 0;
    for (; p; p -= p & -p) ans += a[p];
    return ans;
  }
} b1, b2;
long long a[maxn], w[maxn];
int main() {
  init();
  int n = readint(), q = readint();
  for (int i = 1; i <= n; i++) a[i] = readint() - i;
  b1.n = b2.n = n;
  for (int i = 1; i <= n; i++) {
    w[i] = readint();
    b1.add(i, w[i]);
    b2.add(i, w[i] * a[i] % mod);
  }
  while (q--) {
    int x = readint(), y = readint();
    if (x < 0) {
      x = -x;
      b1.add(x, y - w[x]);
      b2.add(x, y * a[x] % mod - w[x] * a[x] % mod);
      w[x] = y;
      continue;
    }
    long long o = b1.query(x - 1), t = b1.query(y);
    int l = x, r = y;
    while (l < r) {
      int mid = (l + r) / 2;
      if (2 * b1.query(mid) - o - t > 0)
        r = mid;
      else
        l = mid + 1;
    }
    long long ans = (b1.query(l) - o) % mod * a[l] - b2.query(l) +
                    b2.query(x - 1) + b2.query(y) - b2.query(l) -
                    (t - b1.query(l)) % mod * a[l];
    printf("%lld\n", (ans % mod + mod) % mod);
  }
}