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;
namespace mine {
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3fll;
long long qread() {
  long long ans = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' or c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c and c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans * f;
}
void write(long long num) {
  if (num < 0) {
    num = -num;
    putchar('-');
  }
  if (num > 9) write(num / 10);
  putchar('0' + num % 10);
}
void write1(long long num) {
  write(num);
  putchar(' ');
}
void write2(long long num) {
  write(num);
  puts("");
}
void chmax(long long &x, const long long y) { x = x > y ? x : y; }
void chmin(long long &x, const long long y) { x = x < y ? x : y; }
const int MAX_N = 2e5 + 10;
const long long MOD = 1e9 + 7;
void add(long long &x, long long y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
struct BIT {
  long long bit[MAX_N];
  BIT() { memset(bit, 0, sizeof bit); }
  int lowbit(int x) { return x & -x; }
  void add(int x, int c) {
    while (x < MAX_N) bit[x] += c, x += lowbit(x);
  }
  long long sum(int x) {
    long long ans = 0;
    while (x >= 1) ans += bit[x], x -= lowbit(x);
    return ans;
  }
  long long ask(int l, int r) { return (l > r) ? 0 : sum(r) - sum(l - 1); }
} bit, bit2;
long long a[MAX_N], b[MAX_N], w[MAX_N];
void main() {
  int n = qread(), q = qread();
  for (int i = 1; i <= n; i++) a[i] = qread(), b[i] = a[i] - i;
  for (int i = 1; i <= n; i++) {
    w[i] = qread(), bit.add(i, w[i]);
    bit2.add(i, w[i] * b[i] % MOD);
  }
  while (q--) {
    int x = qread(), y = qread();
    if (x < 0) {
      int id = -x, c = y;
      bit.add(id, -w[id]);
      bit2.add(id, -w[id] * b[id] % MOD);
      w[id] = c;
      bit.add(id, w[id]);
      bit2.add(id, w[id] * b[id] % MOD);
    } else {
      int l = x, r = y;
      long long now = 0;
      int T0 = a[l] - l;
      int mid = lower_bound(b + l, b + r + 1, T0) - b - 1;
      now -= bit2.ask(l, mid) % MOD - bit.ask(l, mid) % MOD * T0 % MOD;
      now += bit2.ask(mid + 1, r) % MOD - bit.ask(mid + 1, r) % MOD * T0 % MOD;
      now %= MOD;
      int fl = b[l], fr = b[r] - 1, T = -1;
      while (fl <= fr) {
        int mid = (fl + fr) >> 1;
        int tt = upper_bound(b + 1, b + n + 1, mid) - b - 1;
        if (bit.ask(l, tt) < bit.ask(tt + 1, r))
          T = mid, fl = mid + 1;
        else
          fr = mid - 1;
      }
      if (T >= 0) {
        int tt = upper_bound(b + 1, b + n + 1, T) - b - 1;
        now += -bit2.ask(l, tt) * 2 % MOD +
               bit.ask(l, tt) % MOD * (T + b[l] + 1) % MOD;
        now -= bit.ask(tt + 1, r) % MOD * (T - b[l] + 1) % MOD;
      }
      write2((now % MOD + MOD) % MOD);
    }
  }
}
};  // namespace mine
int main() {
  srand(time(0));
  mine::main();
}