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;
template <class T>
void read(T &x) {
  x = 0;
  char c = getchar();
  long long f = 0;
  while (c < '0' || c > '9') f |= (c == '-'), c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (f) x = -x;
}
const long long N = 2e5 + 5, mod = 1e9 + 7;
void addn(long long &x, long long y) { x = (x + y) % mod; }
long long n, q, a[N], w[N];
struct BIT {
  long long tr[N];
  void init() { memset(tr, 0, sizeof(tr)); }
  void mdf(long long x, long long v, long long op) {
    for (; x <= n; x += x & -x) {
      if (op)
        addn(tr[x], v);
      else
        tr[x] += v;
    }
  }
  long long que(long long x, long long op) {
    long long res = 0;
    for (; x; x -= x & -x) {
      if (op)
        addn(res, tr[x]);
      else
        res += tr[x];
    }
    return res;
  }
  long long ask(long long l, long long r, long long op) {
    if (op)
      return (que(r, 1) - que(l - 1, 1) + mod) % mod;
    else
      return que(r, 0) - que(l - 1, 0);
  }
} sum, mul1, mul2;
signed main() {
  cin >> 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]);
  sum.init();
  mul1.init();
  mul2.init();
  for (long long i = 1; i <= n; ++i) {
    sum.mdf(i, w[i], 0);
    mul1.mdf(i, a[i] * w[i] % mod, 1);
    mul2.mdf(i, w[i] * i % mod, 1);
  }
  long long x, y;
  while (q-- > 0) {
    read(x);
    read(y);
    if (x < 0) {
      x *= -1;
      sum.mdf(x, y - w[x], 0);
      mul1.mdf(x, (a[x] * y % mod - a[x] * w[x] % mod + mod) % mod, 1);
      mul2.mdf(x, (x * y % mod - x * w[x] % mod + mod) % mod, 1);
      w[x] = y;
    } else {
      long long tot = sum.ask(x, y, 0);
      long long l = x, r = y, p, mid;
      while (l <= r) {
        mid = l + r >> 1;
        long long L = sum.ask(x, mid, 0);
        if (L >= tot - L)
          p = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      long long L = 0, R = 0;
      L = (a[p] - p) * sum.ask(x, p - 1, 1) % mod - mul1.ask(x, p - 1, 1) +
          mul2.ask(x, p - 1, 1);
      L = (L % mod + mod) % mod;
      R = mul1.ask(p + 1, y, 1) - (a[p] - p) * sum.ask(p + 1, y, 1) % mod -
          mul2.ask(p + 1, y, 1);
      R = (R % mod + mod) % mod;
      cout << (L + R) % mod << "\n";
    }
  }
  return 0;
}