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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline long long read() {
  long long t = 0, dp = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') dp = -1;
    c = getchar();
  }
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  return t * dp;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
long long mo = 1e9 + 7;
const int N = 200005;
long long n, que, a[N], w[N], tr1[N], tr2[N];
inline long long Ask1(long long x) {
  long long sum = 0;
  for (; x; x -= x & -x) sum = (sum + tr1[x] + mo) % mo;
  return sum;
}
inline long long Ask2(long long x) {
  long long sum = 0;
  for (; x; x -= x & -x) sum += tr2[x];
  return sum;
}
inline void Insert1(long long x, long long v) {
  for (; x <= n; x += x & -x) tr1[x] = (tr1[x] + v % mo + mo) % mo;
}
inline void Insert2(long long x, long long v) {
  for (; x <= n; x += x & -x) tr2[x] += v;
}
inline long long Query1(long long l, long long r) {
  return (Ask1(r) - Ask1(l - 1) + mo) % mo;
}
inline long long Query2(long long l, long long r) {
  return Ask2(r) - Ask2(l - 1);
}
inline bool check(long long l, long long r, long long mid) {
  long long t1 = Query2(l, mid), t2 = Query2(mid + 1, r);
  return t1 >= t2;
}
int main() {
  n = read();
  que = read();
  for (int i = 1; i <= n; ++i) a[i] = read() - i + 1;
  for (int i = 1; i <= n; ++i) w[i] = read();
  for (int i = 1; i <= n; ++i) Insert1(i, a[i] * w[i] % mo), Insert2(i, w[i]);
  for (int i = 1; i <= que; ++i) {
    long long ql = read(), qr = read();
    if (ql > 0) {
      long long l = ql, r = qr, ans = r;
      while (l <= r) {
        long long mid = l + r >> 1;
        if (check(ql, qr, mid))
          ans = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      long long ANS =
          (Query2(ql, ans) % mo * a[ans] % mo - Query1(ql, ans)) % mo;
      ANS +=
          (-Query2(ans + 1, qr) % mo * a[ans] % mo + Query1(ans + 1, qr) % mo) %
          mo;
      ANS = (ANS % mo + mo) % mo;
      writeln(ANS);
    } else {
      ql = -ql;
      Insert1(ql, a[ql] * (qr - w[ql]) % mo);
      Insert2(ql, (qr - w[ql]));
      w[ql] = qr;
    }
  }
}