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 long long P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
const long long N = 1e6 + 10;
long long n, q, pos;
long long a[N], w[N];
long long s1[N << 2], s2[N << 2], sum;
void pu(long long o) {
  s1[o] = s1[(o << 1)] + s1[((o << 1) | 1)];
  s2[o] = (s2[(o << 1)] + s2[((o << 1) | 1)]) % P;
}
void update(long long o, long long l, long long r, long long first,
            long long v) {
  if (l == r) {
    s1[o] = v;
    s2[o] = (long long)v * (a[l] - l) % P;
    return;
  }
  if (((l + r) >> 1) >= first)
    update((o << 1), l, ((l + r) >> 1), first, v);
  else
    update(((o << 1) | 1), ((l + r) >> 1) + 1, r, first, v);
  pu(o);
}
long long qry1(long long o, long long l, long long r, long long ql,
               long long qr) {
  if (ql <= l && r <= qr) return s1[o];
  long long ans = 0;
  if (((l + r) >> 1) >= ql) ans += qry1((o << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) < qr)
    ans += qry1(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
  return ans;
}
long long qry2(long long o, long long l, long long r, long long ql,
               long long qr) {
  if (ql <= l && r <= qr) return s2[o];
  long long ans = 0;
  if (((l + r) >> 1) >= ql) ans = qry2((o << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) < qr)
    (ans += qry2(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr)) %= P;
  return ans;
}
void find(long long o, long long l, long long r, long long ql, long long qr) {
  if (ql <= l && r <= qr) {
    if (l == r || sum == s1[o]) pos = r;
    if (sum >= s1[o])
      sum -= s1[o];
    else
      find((o << 1), l, ((l + r) >> 1), ql, qr);
    return;
  }
  if (((l + r) >> 1) >= ql) find((o << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) < qr) find(((o << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
}
void build(long long o, long long l, long long r) {
  if (l == r) {
    s1[o] = w[l], s2[o] = (long long)(a[l] - l) * w[l] % P;
    return;
  }
  build((o << 1), l, ((l + r) >> 1)),
      build(((o << 1) | 1), ((l + r) >> 1) + 1, r);
  pu(o);
}
long long qry(long long l, long long r) {
  long long L = qry2(1, 1, n, l, r) % P, R = qry1(1, 1, n, l, r) % P;
  (L += P) %= P, (R += P) %= P;
  return ((L + R * (pos - a[pos]) % P) % P + P) % P;
}
signed main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; ++i) scanf("%lld", a + i);
  for (long long i = 1; i <= n; ++i) scanf("%lld", w + i);
  build(1, 1, n);
  for (long long i = 1; i <= q; ++i) {
    long long first, second;
    scanf("%lld%lld", &first, &second);
    if (first < 0)
      update(1, 1, n, -first, second);
    else {
      long long sum = qry1(1, 1, n, first, second);
      long long l = first, r = second, ans = -1;
      while (l <= r) {
        if (2 * qry1(1, 1, n, first, ((l + r) >> 1)) >= sum)
          ans = ((l + r) >> 1), r = ((l + r) >> 1) - 1;
        else
          l = ((l + r) >> 1) + 1;
      }
      pos = ans;
      printf("%lld\n", ((qry(ans, second) - qry(first, ans)) % P + P) % P);
    }
  }
}