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 = 2e5 + 5;
const int mod = 1e9 + 7;
int n, l, r;
long long S[maxn << 1], wp[maxn << 1], owp[maxn], os[maxn], ps[maxn];
long long add(long long a, long long b) {
  if (a + b > mod)
    return a + b - mod;
  else
    return a + b;
}
int lowbit(int x) { return x & (-x); }
void change1(int p, int x) {
  while (p <= n) {
    S[p] += x;
    p += lowbit(p);
  }
}
void change2(int p, int x) {
  while (p <= n) {
    wp[p] = add(x, wp[p]);
    p += lowbit(p);
  }
}
long long ssum(int k) {
  long long ans = 0;
  while (k > 0) {
    ans += S[k];
    k -= lowbit(k);
  }
  return ans;
}
int wpsum(int k) {
  long long ans = 0;
  while (k > 0) {
    ans = add(ans, wp[k]);
    k -= lowbit(k);
  }
  if (ans > mod || ans < -mod)
    return ans % mod;
  else
    return ans;
}
long long ask(int l, int r, int op) {
  if (op == 1)
    return ssum(r) - ssum(l - 1);
  else
    return wpsum(r) - wpsum(l - 1);
}
int findk(int l, int r) {
  int L = l, R = r, mid;
  double al = 1.0 * (ask(l, r, 1)) / 2;
  while (L < R) {
    mid = (L + R) >> 1;
    if (ask(l, mid, 1) >= al)
      R = mid;
    else
      L = mid + 1;
  }
  return L;
}
int main() {
  int q, a, b, pos;
  long long nw, n1, n2;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &ps[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &os[i]);
    owp[i] = (ps[i] - i) * os[i];
    if (owp[i] > mod) owp[i] %= mod;
    change1(i, os[i]);
    change2(i, owp[i]);
  }
  while (q--) {
    scanf("%d%d", &a, &b);
    if (a < 0) {
      pos = -a, nw = b;
      n1 = nw - os[pos], n2 = (nw * (ps[pos] - pos) - owp[pos]) % mod;
      os[pos] = b;
      owp[pos] = nw * (ps[pos] - pos);
      change1(pos, n1);
      change2(pos, n2);
    } else {
      int k = findk(a, b);
      l = a, r = b;
      long long ans = ((ask(l, k, 1) - ask(k, r, 1)) * (ps[k] - k) -
                       (ask(l, k, 2) - ask(k, r, 2))) %
                      mod;
      printf("%lld\n", (ans % mod + mod) % mod);
    }
  }
  return 0;
}