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 mod = 1000 * 1000 * 1000 + 7;
const long long nax = 200005;
struct fenwick_tree_1D {
  long long n, log_n;
  vector<long long> tree;
  fenwick_tree_1D(long long _n)
      : n(_n), log_n((long long)(log2(_n) + 1.0L)), tree(_n + 5) {}
  void update(long long x, long long v) {
    while (x <= n) {
      tree[x] += v;
      x += x & (-x);
    }
  }
  long long prefix_sum(long long x) {
    long long r = 0;
    while (x) {
      r += tree[x];
      x -= x & (-x);
    }
    return r;
  }
  long long range_sum(long long l, long long r) {
    if (l > r) return 0;
    return prefix_sum(r) - prefix_sum(l - 1);
  }
  long long lower_bound(long long v) {
    long long b = 0, s = 0;
    for (long long i = log_n; i >= 0; i--)
      if ((b | (1LL << i)) <= n && s + tree[b | (1LL << i)] < v)
        s += tree[b | (1LL << i)], b |= (1LL << i);
    return (b != n) ? (b + 1) : (1LL << 62);
  }
};
fenwick_tree_1D weights(nax), prefix(nax), suffix(nax);
long long n, q;
long long a[nax] = {0}, w[nax] = {0}, p[nax] = {0}, s[nax] = {0};
void update(long long x, long long y) {
  weights.update(x, y - w[x]);
  w[x] = y;
  prefix.update(x, ((a[x] + n - x + 1) * y - p[x] + mod) % mod);
  p[x] = ((a[x] + n - x + 1) * y) % mod;
  suffix.update(x, ((a[x] + x) * y - s[x] + mod) % mod);
  s[x] = ((a[x] + x) * y) % mod;
}
long long query(long long l, long long r) {
  long long totalWeight = weights.range_sum(l, r);
  long long median = weights.prefix_sum(l - 1) + (totalWeight + 1) / 2;
  long long medianPosition = weights.lower_bound(median);
  long long leftWeight = weights.range_sum(l, medianPosition) % mod;
  long long rightWeight = weights.range_sum(medianPosition + 1, r) % mod;
  long long leftSum =
      (((a[medianPosition] + n - medianPosition + 1) * leftWeight) % mod) -
      (prefix.range_sum(l, medianPosition) % mod);
  leftSum += mod;
  leftSum %= mod;
  long long rightSum =
      (prefix.range_sum(medianPosition + 1, r) % mod) -
      (((a[medianPosition] + n - medianPosition + 1) * rightWeight) % mod);
  rightSum += mod;
  rightSum %= mod;
  return (leftSum + rightSum) % mod;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    update(i, x);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0)
      update(-1 * x, y);
    else
      cout << query(x, y) << "\n";
  }
}