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;
long long a[(long long)1e6], w[(long long)1e6];
const long long maxn = 100100;
const long long mod = 1e9 + 7;
long long tree[maxn << 2], tree2[maxn << 2], tree3[maxn << 2];
long long n;
inline void add(long long r, long long v) {
  while (r <= n) tree[r] += v, r += r & -r;
}
inline long long query(long long r) {
  long long ans = 0;
  while (r) ans += tree[r], r &= r - 1;
  return ans;
}
inline long long query(long long l, long long r) {
  return query(r) - query(l - 1);
}
inline void add1(long long r, long long v) {
  while (r <= n) (tree2[r] += v) %= mod, r += r & -r;
}
inline long long query1(long long r) {
  long long ans = 0;
  while (r) (ans += tree2[r]) %= mod, r &= r - 1;
  return ans;
}
inline long long query1(long long l, long long r) {
  return query1(r) - query1(l - 1);
}
inline void add2(long long r, long long v) {
  while (r <= n) (tree3[r] += v) %= mod, r += r & -r;
}
inline long long query2(long long r) {
  long long ans = 0;
  while (r) (ans += tree3[r]) %= mod, r &= r - 1;
  return ans;
}
inline long long query2(long long l, long long r) {
  return query2(r) - query2(l - 1);
}
inline long long move_1(long long l, long long r, long long pos) {
  long long len = r - l + 1;
  return pos * (query(l, r) % mod) % mod + query2(l, r) -
         query(l, r) * l % mod - query1(l, r);
}
inline long long move_2(long long l, long long r, long long pos) {
  long long len = r - l + 1;
  return -pos * (query(l, r) % mod) % mod - query2(l, r) +
         query(l, r) % mod * l % mod + query1(l, r);
}
signed main() {
  ios::sync_with_stdio(false);
  long long q;
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; ++i) {
    cin >> w[i];
    add(i, w[i]);
    add1(i, a[i] * w[i] % mod);
    add2(i, (i)*w[i] % mod);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      add(x, -w[x]);
      add1(x, mod - a[x] * w[x] % mod);
      add2(x, mod - (x)*w[x] % mod);
      w[x] = y;
      add(x, w[x]);
      add1(x, a[x] * w[x] % mod);
      add2(x, (x)*w[x] % mod);
    } else {
      long long l = x, r = y;
      while (l != r) {
        long long mid = l + r >> 1;
        if (query(x, mid) < query(mid + 1, y))
          l = mid + 1;
        else
          r = mid;
      }
      cout << ((move_1(x, r, a[r] - (r - x + 1) + 1) +
                move_2(r + 1, y, a[r] + 1)) %
                   mod +
               mod) %
                  mod
           << endl;
    }
  }
}