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 GCC optimize "-O3"
#pragma GCC target("avx,avx2,fma")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
int n;
int q;
vector<long long int> v;
vector<long long int> w;
long long int bit[300002];
void add(int i, long long int x) {
  i++;
  while (i < 300002) {
    bit[i] += x;
    i += i & -i;
  }
}
long long int sum(int i) {
  i++;
  long long int r = 0;
  while (i) {
    r += bit[i];
    i -= i & -i;
  }
  return r;
}
long long int rng(int l, int r) {
  long long int sm = sum(r);
  if (l) sm -= sum(l - 1);
  return sm;
}
long long int pf[300002];
void pf_add(int i, long long int x) {
  x %= 1000000007;
  x += 1000000007;
  x %= 1000000007;
  i++;
  while (i < 300002) {
    pf[i] += x;
    if (pf[i] >= 1000000007) pf[i] %= 1000000007;
    i += i & -i;
  }
}
long long int pf_sum(int i) {
  i++;
  long long int r = 0;
  while (i) {
    r += pf[i];
    if (r >= 1000000007) r %= 1000000007;
    i -= i & -i;
  }
  return r;
}
long long int sf[300002];
void sf_add(int i, long long int x) {
  i++;
  x %= 1000000007;
  x += 1000000007;
  x %= 1000000007;
  while (i < 300002) {
    sf[i] += x;
    if (sf[i] >= 1000000007) sf[i] %= 1000000007;
    i += i & -i;
  }
}
long long int sf_sum(int i) {
  i++;
  long long int r = 0;
  while (i) {
    r += sf[i];
    if (r >= 1000000007) r %= 1000000007;
    i -= i & -i;
  }
  return r;
}
int main() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    v.push_back(a);
  }
  int rngg = n;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    w.push_back(a);
    add(i, a);
    pf_add(i, -(v[i] + rngg) * a);
    sf_add(i, (v[i] - i) * a);
    rngg--;
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x *= -1;
      x--;
      add(x, -w[x]);
      pf_add(x, (v[x] + (n - x)) * w[x]);
      sf_add(x, -(v[x] - x) * w[x]);
      w[x] = y;
      add(x, w[x]);
      pf_add(x, -(v[x] + (n - x)) * w[x]);
      sf_add(x, (v[x] - x) * w[x]);
    } else {
      int l = x;
      int r = y;
      l--;
      r--;
      int mint = l;
      int maxt = r;
      long long int ALL = (rng(l, r) + 1LL) / 2LL;
      while (mint + 1 < maxt) {
        int mid = (mint + maxt) / 2;
        if (rng(l, mid) >= ALL) {
          maxt = mid;
        } else {
          mint = mid;
        }
      }
      if (rng(l, mint) >= ALL) {
        maxt = mint;
      } else {
        mint = maxt;
      }
      long long int act = n - mint;
      long long int cs1 = (((sum(mint) - sum(l - 1)) % 1000000007) *
                           (v[mint] + act) % 1000000007) +
                          ((pf_sum(mint) - pf_sum(l - 1)) % 1000000007);
      act = mint;
      cs1 %= 1000000007;
      long long int cs2 = ((sf_sum(r) - sf_sum(mint - 1) % 1000000007)) -
                          (((sum(r) - sum(mint - 1)) % 1000000007) *
                           (v[mint] - act) % 1000000007);
      cs2 %= 1000000007;
      cs1 += cs2;
      cs1 %= 1000000007;
      cs1 += 1000000007;
      cs1 %= 1000000007;
      printf("%lld\n", cs1);
    }
  }
  return 0;
}