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 MAXN = 200100;
const int P = (1 << 18);
const int MAXP = 2 * P + 10;
const int MOD = 1e9 + 7;
int N, Q;
long long a[MAXN];
long long w[MAXN];
long long wseg[MAXP], wiseg[MAXP], waseg[MAXP];
void upd(long long* seg, int cloc, long long cdiff, bool mmod) {
  cloc += P;
  while (cloc) {
    seg[cloc] += cdiff;
    if (mmod) seg[cloc] %= MOD;
    cloc /= 2;
  }
}
long long add(long long* seg, int cloc, bool mmod) {
  long long res = 0;
  cloc += P;
  while (cloc > 1) {
    if (cloc & 1) {
      res += seg[cloc - 1];
      if (mmod) res %= MOD;
    }
    cloc /= 2;
  }
  return res;
}
int wsearch(long long sgoal) {
  int cloc = 1;
  while (cloc < P) {
    if (wseg[2 * cloc] >= sgoal)
      cloc *= 2;
    else {
      sgoal -= wseg[2 * cloc];
      cloc *= 2;
      cloc++;
    }
  }
  return cloc - P;
}
int findm(int lo, int hi) {
  long long sleft = add(wseg, lo, 0), sright = add(wseg, hi + 1, 0);
  long long sgoal = (sleft + sright + 1) / 2;
  return wsearch(sgoal);
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 0; i < N; i++) {
    int x;
    scanf("%d", &x);
    a[i] = x;
  }
  for (int i = 0; i < N; i++) {
    int x;
    scanf("%d", &x);
    w[i] = x;
  }
  for (int i = 0; i < N; i++) {
    upd(wseg, i, w[i], 0);
    upd(wiseg, i, i * w[i], 1);
    upd(waseg, i, a[i] * w[i], 1);
  }
  for (int i = 0; i < Q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    if (l < 0) {
      int cloc = -l - 1;
      long long cdiff = r - w[cloc];
      upd(wseg, cloc, cdiff, 0);
      upd(wiseg, cloc, cloc * cdiff, 1);
      upd(waseg, cloc, a[cloc] * cdiff, 1);
      w[cloc] = r;
    } else {
      l--, r--;
      int mm = findm(l, r);
      long long ws = ((add(wseg, mm, 1) - add(wseg, l, 1)) % MOD + MOD) % MOD;
      long long wis =
          ((add(wiseg, mm, 1) - add(wiseg, l, 1)) % MOD + MOD) % MOD;
      long long was =
          ((add(waseg, mm, 1) - add(waseg, l, 1)) % MOD + MOD) % MOD;
      long long res = ws * (a[mm] - mm) + wis - was;
      r++;
      ws = ((add(wseg, r, 1) - add(wseg, mm, 1)) % MOD + MOD) % MOD;
      wis = ((add(wiseg, r, 1) - add(wiseg, mm, 1)) % MOD + MOD) % MOD;
      was = ((add(waseg, r, 1) - add(waseg, mm, 1)) % MOD + MOD) % MOD;
      long long bres = ws * (mm - a[mm]) - wis + was;
      res = ((res + bres) % MOD + MOD) % MOD;
      printf("%d\n", (int)res);
    }
  }
}